Question

A boy throws a ball with a velocity of 20 ms-'. Find the time elapsed
between the throwing and catching the ball?​

Answer 1

Answer:

If he’s throwing the ball straight up, then the equation for the ball’s motion is h(t)=-4.9 m/s²*t²+20 m/s * t + 0 ( assuming the boy is on the ground when he throws it). The time elapsed between throwing and landing is given by when h(t)=0. So:

h(t)=-4.9t²+20t+0

0=-4.9t²+20t

20t=4.9t²

4.9t=20

t=20/4.9= 4.082 secs

If the air resistance is assumed to be negligible during the upward and downward journey of the ball, then the total time elapsed between the throwing and catching the ball will be 4.081 seconds.

The solution is attached in the image.

Thanks!!

Answer 2

Answer:

f he’s throwing the ball straight up, then the equation for the ball’s motion is h(t)=-4.9 m/s²*t²+20 m/s * t + 0 ( assuming the boy is on the ground when he throws it). The time elapsed between throwing and landing is given by when h(t)=0

Explanation:

h(t)=-4.9t²+20t+0

0=-4.9t²+20t

20t=4.9t²

4.9t=20

t=20/4.9= 4.082 secs