Question

a ball thrown up is caught by the thrower after 4s. With what velocity was it thrown up? how high did it go? where was it after 3s?(g=9.8ms-square)​

Answer 1

Explanation:

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Answer 2

Answer:

If a ball that was thrown up is caught by the thrower after 4 seconds, then how far below its highest point was it at 3 seconds from the start?

4 second here indicates the total time taken by the ball to reach highest point and come back to the thrower. So the highest point is reached at half the time i.e. 2 seconds. take g=9.8m/s^2

v=u-gt (since ball goes upward)

at highest point v=0, time taken=2

u=v+gt

u=19.6m/s…….

v^2=u^2-2gh (ball goes upward)

at highest point v=0

h=u^2/2g

h=19.6m……..(highest point)

h=ut- .5gt^2

t=3 seconds, u=19.6

h=14.7m

so your answer would be= 19.6–14.7= 4.9m

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