Question

A farm truck moves due east with a constant velocity of 9.50 m/s on a limitless, horizontal stretch of road. A boy riding on the back of the truck throws a can of soda upward (Fig. P4.54) and catches the projectile at the same location on the truck bed, but 16.0 m farther down the road. (a) In the frame of reference of the truck, at what angle to the vertical does the boy throw the can? (b) What is the initial speed of the can relative to the truck?

Answer 1

Answer

given,

constant velocity = 9.50 m/s

acceleration = 0

distance = 16 m

the time of flight

$s = s_o + u t + \dfrac{1}{2}at^2$

16 = 9.50 × t

t = 1.68 s

a) to catch the can the truck he must throw it  straight up, at 0° to the vertical.

b) now,

$y_f = Y_i + v_yt + \dfrac{1}{2}at^2$

$0 = 0 +  v_y\times 1.68 - \dfrac{1}{2}\times 9.8 \times 1.68^2$

$v_y = 8.23 m/s$