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AP Physics C: Mechanics : Calculating Motion in Two Dimensions

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Example Questions

Example Question #1 : Calculating Motion In Two Dimensions

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Possible Answers:

I and II

I, II, and III

I only

I and III

II only

Correct answer:

I and III

Explanation:

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range R_x , so increasing $v_{ox}$ increases R_x if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon a_y would be less. Decreasing a_y increases the time the ball is in the air, thereby increasing R_x if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

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Example Question #1 : Calculating Motion In Two Dimensions

Water emerges horizontally from a hole in a tank $1.5\:m$ above the ground. If the water hits the ground $2.1\: m$ from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Possible Answers:

$3.8\:\uptext{m/s}$

$5.13\:\uptext{m/s}$

$4.88\:\uptext{m/s}$

$4.34\:\uptext{m/s}$

$5.60\:\uptext{m/s}$

Correct answer:

$3.8\:\uptext{m/s}$

Explanation:

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is $9.8\:m/s^2$ and $d_y=1.5\:m$ .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\\ \\ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55\:\uptext{s}$

Next we have to substitute the time into the equation for the x component

d_x = v_x t

We know that $dx = 2.1\:m$ and $t= 0.55\:s$ , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80\:m/s$

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AP Physics C: Mechanics : Calculating Motion in Two Dimensions

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 : Calculating Motion In Two Dimensions

Example Question #1 : Calculating Motion In Two Dimensions

All AP Physics C: Mechanics Resources

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