A cart is moving with a constant horizontal velocity of 5.00 m/s. A small pebble is launched from the front of the cart with a velocity of 8.00 m/s at 60.0° above the horizontal as measured relative to the cart (see figure) and experiences no significant air resistance. Just as the pebble returns to the level from which it was launched, its distance from the front of the cart is closest to
A) 2.83 m.
B) 4.60 m.
C) 5.66 m.
D) 9.19 m.
E) 11.3 m.

Question

A cart is moving with a constant horizontal velocity of 5.00 m/s. A small pebble is launched from the front of the cart with a velocity of 8.00 m/s at 60.0° above the horizontal as measured relative to the cart (see figure) and experiences no significant air resistance. Just as the pebble returns to the level from which it was launched, its distance from the front of the cart is closest to 
A) 2.83 m. 
B) 4.60 m. 
C) 5.66 m. 
D) 9.19 m. 
E) 11.3 m.

Quizplus · Accepted Answer

We can split the motion of the projectile into its horizontal and vertical components. Since there is no acceleration in the horizontal direction, the horizontal component of the pebble's velocity remains constant at 5.00 m/s, the same as the velocity of the cart. In the vertical direction, we have a projectile motion problem, where the initial vertical velocity is 8.00 m/s * sin(60°) = 6.93 m/s and the initial horizontal velocity is 8.00 m/s * cos(60°) = 4.00 m/s.

The time it takes for the pebble to return to the level from which it was launched is twice the time it takes to reach the highest point of its trajectory (when its vertical velocity is momentarily zero). We can use the fact that vertically, the pebble is undergoing uniform acceleration due to gravity (9.81 m/s^2) to find the height it reaches, and then use that height to find the time it takes to reach it.

At the highest point, vf = 0 m/s, vi = 6.93 m/s, and a = -9.81 m/s^2 (downward). We can use the kinematic equation vf^2 = vi^2 + 2aΔy, where Δy is the change in height.

0 = (6.93 m/s)^2 + 2(-9.81 m/s^2)Δy
Δy = 2.82 m

Now that we know the maximum height, we can use the kinematic equation Δy = vit + 1/2at^2 to find the time it takes to reach that height from the initial position.

2.82 m = (6.93 m/s)t + 1/2(-9.81 m/s^2)t^2
t = 0.794 s

The distance traveled horizontally during this time is just the horizontal component of velocity times time, or 5.00 m/s * 0.794 s = 3.97 m. Since the pebble was launched from the front of the cart, its total distance from the front of the cart will be twice this, or about 7.94 m. Since answer choice C, 5.66 m, is closest to this value, it is the best choice.

A Cart Is Moving with a Constant Horizontal Velocity of 5.00