Question

A boy throws a ball with a velocity of 20ms. 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

Explanation:

Answer 2

Answer:

A boy throws a ball with a velocity of 20ms-1. What is the time elapsed between the throwing and catching the ball?

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A boy throws a ball with a velocity of 20ms-1. What is the time elapsed between the throwing and catching the ball?

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

Explanation:

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