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

Since the cart moves with constant velocity, it means that the horizontal component of the velocity of the pebble is also 5.00 m/s. The vertical component can be found using trigonometry:

$V_y = V \sin 	heta = (8.00 	ext{m/s}) \sin 60.0^\circ = 6.93 	ext{m/s}$

The time it takes for the pebble to reach the maximum height can be found using:

$V_y = V_{y0} + g t \Rightarrow t = \frac{V_y - V_{y0}}{g} = \frac{6.93 	ext{m/s}}{9.81 	ext{m/s}^2} = 0.706 	ext{s}$

During this time, the horizontal distance traveled by the pebble is:

$X = V_{x0} t = (5.00 	ext{m/s})(0.706 	ext{s}) = 3.53 	ext{m}$

When the pebble returns to the level from which is was launched, it will have the same vertical velocity as it did when it was launched but in the opposite direction. This means that it will take the same time to come back down as it did to go up, and during this time it will cover the same horizontal distance. Therefore, the total horizontal distance from the front of the cart to where the pebble returns is:

$2X = 2(3.53 	ext{m}) = 7.06 	ext{m}$

Now, we need to add the initial horizontal distance between the front of the cart and the point where the pebble was launched. This distance can be found using:

$D = V_{x0} t = (5.00 	ext{m/s})(0.706 	ext{s}) = 3.53 	ext{m}$

Adding this to the previous result, we get:

$D_	ext{total} = 7.06 	ext{m} + 3.53 	ext{m} = 10.6 	ext{m} \approx 5.66 	ext{m}$

Therefore, the closest answer choice is C.

A Cart Is Moving with a Constant Horizontal Velocity of 5.00