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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

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6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Linear Motion

If you dropped a penny from the roof of a building 15m high, how much time would elapse before the penny hits the ground?

$g=10\frac{m}{s^2}$

Possible Answers:

t=1.93s

t=3.73s

t=2.2s

t=2.3s

t=1.73s

Correct answer:

t=1.73s

Explanation:

Use one of the kinematic equations. The best way to select the correct one is by process of elimination based on which variables we have or don't. The problem gives us vertical displacement, gravity, and if we look closely, initial velocity (since the stone was "dropped" it equals 0). Based on these we will definitely use one that factors in displacement:

$\Delta y=V_{0}t+.5at^2$

Note that these kinematic equations normally have an x-variable for displacement, but since our problem has an object falling vertically we use y-variables for consistency.

Plug in given values and solve:

$15={\color{Red} V_{0}t}+.5(10)t^2$

The red term becomes 0 since the initial velocity is 0.

Simplify and solve for t:

15=5t^2

3=t^2

t=1.73s

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Example Question #31 : Linear Motion

You are driving at $25\frac{m}{s}$ when you slam on the brakes and skid to a halt just before the red light over the course of 2s. How far did the car slide? Assume uniform acceleration.

Possible Answers:

$\Delta x=30m$

$\Delta x=19m$

None of these

$\Delta x=18m$

$\Delta x=25m$

Correct answer:

$\Delta x=25m$

Explanation:

We know the initial velocity, the final velocity, and the time it took for the car to stop moving after the brakes were applied. Given the final and initial velocity we can calculate average velocity. Average velocity multiplied by time is equal to the distance traveled.

$D=\frac{V_{o}+V}{2}*t$

$D=\frac{25+0}{2}*2=25m$

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Example Question #31 : Motion And Mechanics

A flea can jump nearly 100 times its height. If a flea can jump 0.18m vertically, what is its take off speed (speed at which it leaves the ground)?

$g=10\frac{m}{s^2}$

Possible Answers:

$V_{0}=1.8\frac{m}{s}$

$V_{0}=1.9\frac{m}{s}$

$V_{0}=2.6\frac{m}{s}$

$V_{0}=2.4\frac{m}{s}$

$V_0=3.2\frac{m}{s}$

Correct answer:

$V_{0}=1.9\frac{m}{s}$

Explanation:

Since we are given the vertical displacement, acceleration due to gravity, and we know that the final velocity in the y-direction is zero since it will be at the top of its jump. Use the following kinematics formula and solve for the initial velocity.

$V^2=V_{0}^2+2a\Delta y$

Plug in known values and solve for initial velocity.

$0=V_{0}^2+2(-10\frac{m}{s^2}).18m$

$0=V_{0}^2-20\frac{m}{s^2}*.18m$

$0=V_{0}^2-3.6\frac{m^2}{s^2}$

$3.6\frac{m^2}{s^2}=V_{0}^2$

$\sqrt{3.6\frac{m^2}{s^2}}=\sqrt{V_{0}^2}$

$V_{0}=1.9\frac{m}{s}$

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Example Question #34 : Linear Motion

If you threw a tomato upwards with an initial velocity of $12\frac{m}{s}$ , at what time (in seconds) would the tomato hit the ground?

$g=10\frac{m}{s^2}$

Possible Answers:

t=2.2s

t=2.4s

t=3.0s

t=4.3s

t=2.8s

Correct answer:

t=2.4s

Explanation:

Gravity accelerates everything downward by $10\frac{m}{s^2}$ . When the tomato is thrown upward with some velocity, gravity immediately begins to slowly reduce this velocity since the acceleration opposes the direction of the velocity.

$V_{0}=12\frac{m}{s}$ (initial velocity)

$V=0\frac{m}{s}$ (final velocity is equal to 0)

$t=\frac{V-V_{0}}{a}=\frac{0-12\frac{m}{s}}{-10\frac{m}{s^2}}=1.2s$

In 1.2s the tomato will reach its maximum height. If you have ever thrown a ball upward you may have noticed how it appears to stop at the peak. We have just calculated the time it takes for that ball to appear to stop for a very small time and fall back down. Since our tomato must travel back to Earth, we double the time for its up and down motion (since they equal each other) to get 2.4s as the final answer.

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Example Question #1 : Understanding Motion In Two Dimensions

A ball rolls off of a table with an initial horizontal velocity of $19\frac{m}{s}$ . If the table is high, how long will it take to hit the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

0.06s

0.155s

1.28s

0.612s

0.782s

Correct answer:

0.782s

Explanation:

The problem gives us the initial horizontal velocity. This velocity will only affect the distance the ball travels in the horizontal direction; it will have no effect on the time the ball is in the air. Since there is no angle of trajectory, the ball has no initial vertical velocity.

We know the height of the table, the initial velocity, and gravity. Using these values with the appropriate motion equation, we can solve for the time.

The best equation to use is:

$\Delta y =v_it+\frac{1}{2}at^2$

We can use our values to solve for the time. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

$-3m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m=(-4.9\frac{m}{s^2})t^2$

$\frac{-3m}{-4.9\frac{m}{s^2}}=t^2$

0.612s^2=t^2

$\sqrt{0.612s^2}=\sqrt{t^2}$

0.782s=t

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Example Question #2 : Understanding Motion In Two Dimensions

A ball rolls off of a table with an initial horizontal velocity of $19\frac{m}{s}$ . If the table is high, what is the final horizontal velocity?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

$6.33\frac{m}{s}$

$0.782\frac{m}{s}$

$19\frac{m}{s}$

$3\frac{m}{s}$

$1.34\frac{m}{s}$

Correct answer:

$19\frac{m}{s}$

Explanation:

One of the key concepts of parabolic motion in freefall is that the horizontal velocity, v_x , does not change. In order for velocity to change, there must be an acceleration. If an object is accelerating, then there must be a force acting on the object. The only force on the object during freefall is gravity, which acts in the vertical direction and cannot affect the horizontal velocity.

Since our given $v_{ix}=19\frac{m}{s}$ , our $v_{fx}=19\frac{m}{s}$ .

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Example Question #3 : Understanding Motion In Two Dimensions

A ball rolls off of a table with an initial horizontal velocity of $19\frac{m}{s}$ . If the table is high, what is angle the ball will make above the horizontal as it strikes the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

$20.5^\circ$

$30^\circ$

$15^\circ$

$22^\circ$

$68^\circ$

Correct answer:

$22^\circ$

Explanation:

In order for us to find the final angle, we need to find the final horizontal and vertical components.

The only force acting on the ball while in the air is gravity. Since gravity acts in the vertical direction, there is no force in the horizontal direction. This means that the horizontal velocity will not change.

$v_{xi}=v_{xf}=19\frac{m}{s}$

The problem states that the initial velocity is only in the horizontal direction; the initial vertical velocity is zero. We now know initial velocity, acceleration, and distance traveled.

Remember, even though the distance it will travel is , its displacement will be -3m as it moves in the downward direction.

Using these values and the appropriate motion equation, we can solve for the final velocity. The best equation to use is:

$v_{fy}^2=v_{iy}^2+2a\Delta y$

Use the given values to find the final velocity.

$v_{fy}^2=0\frac{m}{s}^2 +2(-9.8\frac{m}{s^2})(-3m)$

$v_{fy}^2=58.8\frac{m^2}{s^2}$

$\sqrt{v_{fy}^2}}=\sqrt{58.8\frac{m^2}{s^2}}$

$v_{fy}=7.67\frac{m}{s}$

Now that we know the horizontal and vertical components of the final velocity, we can solve for the angle by considering the components are legs of a right triangle. Using trigonometric identities, we can solve for the angle.

$\tan(\Theta)=\frac{v_y}{v_x}$

Using our horizontal and vertical velocities, we can calculate the angle.

$\tan(\Theta)=\frac{7.67\frac{m}{s}}{19\frac{m}{s}}$

$\tan(\Theta)=0.404$

$\Theta=\tan^{-1}(0.404)$

$\Theta=22^\circ$

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Example Question #4 : Understanding Motion In Two Dimensions

A ball rolls off of a table with an initial horizontal velocity of $19\frac{m}{s}$ . If the table is high, what will be its final total speed?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

$209.9\frac{m}{s}$

$58.83\frac{m}{s}$

$419.83\frac{m}{s}$

$41\frac{m}{s}$

$20.49\frac{m}{s}$

Correct answer:

$20.49\frac{m}{s}$

Explanation:

In order for us to find the final resultant speed, we need to find the final horizontal and vertical components.

$v_{xi}=v_{xf}=19\frac{m}{s}$

The problem states that the initial velocity is only in the horizontal direction; the initial vertical velocity is zero. We now know initial velocity, acceleration, and distance traveled.

Remember, even though the distance it will travel is , its displacement will be -3m as it moves in the downward direction.

Using these values and the appropriate motion equation, we can solve for the final velocity. The best equation to use is:

$v_{fy}^2=v_{iy}^2+2a\Delta y$

Use the given values to find the final velocity.

$v_{fy}^2=0\frac{m}{s}^2 +2(-9.8\frac{m}{s^2})(-3m)$

$v_{fy}^2=58.8\frac{m^2}{s^2}$

$\sqrt{v_{fy}^2}}=\sqrt{58.8\frac{m^2}{s^2}}$

$v_{fy}=7.67\frac{m}{s}$

Now that we know the final velocity in both the horizontal and vertical directions, we can use the Pythagorean theorem to solve for the total velocity.

$v_{resultant}^2=v_x^2+v_y^2$

$v_{resultant}^2=(19\frac{m}{s})^2+(7.67\frac{m}{s})^2$

$v_{resultant}^2=361\frac{m}{s}+58.83\frac{m}{s}$

$v_{resultant}^2=419.83\frac{m}{s}$

$\sqrt{v_{resultant}^2}=\sqrt{419.83\frac{m}{s}}$

$v_{resultant}=20.49\frac{m}{s}$

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Example Question #41 : Linear Motion

A ball rolls off of a table with an initial horizontal velocity of $19\frac{m}{s}$ . If the table is high, what is the final vertical velocity?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

$3.83\frac{m}{s}$

$-7.67\frac{m}{s}$

$19.6\frac{m}{s}$

$15.34\frac{m}{s}$

$-58.8\frac{m}{s}$

Correct answer:

$-7.67\frac{m}{s}$

Explanation:

The problem states that the initial velocity is only in the horizontal direction; the initial vertical velocity is zero. We now know initial velocity, acceleration, and distance traveled.

Remember, even though the distance it will travel is , its displacement will be -3m as it moves in the downward direction.

Using these values and the appropriate motion equation, we can solve for the final velocity. The best equation to use is:

$v_{fy}^2=v_{iy}^2+2a\Delta y$

Use the given values to find the final velocity.

$v_{fy}^2=0\frac{m}{s}^2 +2(-9.8\frac{m}{s^2})(-3m)$

$v_{fy}^2=58.8\frac{m^2}{s^2}$

$\sqrt{v_{fy}^2}}=\sqrt{58.8\frac{m^2}{s^2}}$

$v_{fy}=7.67\frac{m}{s}$

Because we just took the square root of a number, we got an absolute value for our $v_{fy}$ ; however, velocity is a vector and can be either positive or negative depending on direction. Because the ball is headed downward, the final velocity should correctly be $-7.67\frac{m}{s}$ . Remember that a negative number squared gives a positive value, just like a positive number.

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Example Question #42 : Linear Motion

A ball rolls off of a table with an initial horizontal velocity of $19\frac{m}{s}$ . If the table is high, how far from the table will it land?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

14.86m

0.782m

19m

29.72m

7.43m

Correct answer:

14.86m

Explanation:

We can solve for the horizontal distance using only the horizontal velocity: $\Delta x =v_xt$ .

We are given the value of v_x , but we need to find the time. Time in the air will be determined by the vertical components of the ball's motion.

We know the height of the table, the initial velocity, and gravity. Using these values with the appropriate motion equation, we can solve for the time.

The best equation to use is:

$\Delta y =v_it+\frac{1}{2}at^2$

We can use our values to solve for the time. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

$-3m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m=(-4.9\frac{m}{s^2})t^2$

$\frac{-3m}{-4.9\frac{m}{s^2}}=t^2$

0.612s^2=t^2

$\sqrt{0.612s^2}=\sqrt{t^2}$

0.782s=t

Now we have both the time and the horizontal velocity. Use the original equation to solve for the distance.

$\Delta x =v_xt$

$\Delta x =(19\frac{m}{s})(0.782s)$

$\Delta x =14.86m$

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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #31 : Linear Motion

Example Question #31 : Linear Motion

Example Question #31 : Motion And Mechanics

Example Question #34 : Linear Motion

Example Question #1 : Understanding Motion In Two Dimensions

Example Question #2 : Understanding Motion In Two Dimensions

Example Question #3 : Understanding Motion In Two Dimensions

Example Question #4 : Understanding Motion In Two Dimensions

Example Question #41 : Linear Motion

Example Question #42 : Linear Motion

All High School Physics Resources

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