Question

A ball is thrown straight downward with a speed of 0.50 meter per second from a height of 4.0 meters. What is the speed of the ball 0.70 seconds after it is released?

Answer 1

Assuming that reaching a height 0 doesn’t stop the ball, and that it accelerates at 9.8 m/s^2, the ball would be traveling at 0.5 + 0.7*9.8 = 7.36 m/s downwards.

Answer 2

Answer: The speed of the ball 0.70 seconds after release is 7.36 m/s

     

Explanation:

Using equation of motion in vertical direction and considering downward direction as positive

$v_y=u_y +a_yt$

where $u_y= 0.50 \frac{m}{s} , a_y=g=9.80 \frac{m}{s^{2}} , t=0.70s$

=> $v_y=(0.5+9.8\times 0.70) \frac{m}{s}= 7.36\frac{m}{s}$

Thus the speed of the ball 0.70 seconds after release is 7.36 m/s