Question

a boy throws a ball with a velocity of 20 metre per second find the time elapsed between the throwing and catching the ball ​

Answer 1

Answer:

4.082 secs

Explanation:

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

Answer 2

Answer:

The time elapsed between the throwing and catching of the ball= 4.08 seconds

Explanation:

we will assume that the ball is thrown vertically upward and caught at the same place.

I this case the ball will go upward till it's velocity becomes 0 m/s then it will free fall.

v = u + at

v - final velocity = 0

u - initial velocity = 20

t - time

a- acceleration or retardation due to

gravity

t = (v - u) = a = (0-20) ÷ (- 9.80)

-20

=

- 9.8

= 20 -9.8

= 2.04 seconds

now this will free fall from there

and will take same time to be caught =

2.04 seconds

So the time elapsed between the throwing and catching the ball = 2.04 +2.04 = 4.08 seconds