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 horizontal velocity of the cart and the vertical velocity of the pebble are independent of each other, we can analyze them separately.

Horizontal motion:

The pebble is launched with a horizontal velocity of 5.00 m/s + 8.00cos(60.0°) = 8.00 m/s. Since there is no acceleration in the horizontal direction, the horizontal position of the pebble relative to the cart remains constant throughout its motion.

Vertical motion:

The initial vertical velocity of the pebble is 8.00sin(60.0°) = 6.93 m/s. The time it takes for the pebble to reach the highest point can be found using:

v = u + at

where v = 0 m/s (at highest point), u = 6.93 m/s, a = -9.81 m/s^2 (acceleration due to gravity), and t is the time to reach the highest point. Solving for t, we get:

t = u/a = 0.706 s

The time it takes for the pebble to return to the same level can be found using:

s = ut + 0.5at^2

where s is the vertical distance the pebble travels (in this case, equal to the initial height), all other variables remain the same. Solving for t, we get:

t = sqrt(2s/a)

Substituting in the values we know, we get:

0.706 s = sqrt(2s/-9.81 m/s^2)

Solving for s, we get:

s = 0.5*(-9.81 m/s^2)*(0.706 s)^2 = 2.49 m

Since the horizontal distance traveled by the pebble is equal to the distance traveled by the front of the cart during the same time period (0.706 s), the total distance between the front of the cart and the pebble when the pebble returns to its original level is:

sqrt((8.00 m/s)(0.706 s))^2 + (2.49 m)^2 = 5.66 m

Therefore, the closest answer choice is C) 5.66 m.

A Cart Is Moving with a Constant Horizontal Velocity of 5.00