A ball thrown up is caught by the thrower after 4s. How high did it go and with what velocity was it thrown ?

Explanation

We are given Time of flight = 4 s
We are to calculate : height h and velocity u
We know that ,
time of ascent = time of descent, time of ascent = 2 s
At the highest point, `v =0 `
Using relation: `v = u + g t, ` we get
`0=u -9.8 xx 2 (because g =- 9. 8 ms ^(-2))`
or `u = 19. 6 ms ^(-1)`
We know that , `v ^(2) - u ^(2) = 2 gh`
`therefore 0 ^(2) - ( 19. 6) ^(2) =2 xx (- 9.8) implies h = ( 19. 6 x 19. 6)/( 2 xx 9.8) = 19 .6 m`