Question

A ball is thrown vertically upward from the top of a tower of height h with velocity v the ball strikes the ground after time

Answer 1

this is your answer hope this helped.

Answer 2

Answer:

$Total\ time\ is =\dfrac{v}{g}+\sqrt{\dfrac{2(h+\dfrac{v^2}{2g})}{g}}$

Explanation:

Given that

Height of water = h

Velocity = v

Time taken by ball from A to B is t

We know that acceleration due to gravity is in downward direction

The final speed of ball at point B will be zero.

Vf= Vi + a t

0 = v-  g t

t= v/g

We also know that

$V_f^2=V_i^2+2as$

$0=V^2-2\times g\times d$

d= v²/2g

The total distance from B to D = h + d

The time taken by ball from B to D is T

Initial velocity of ball at B is zero.

We know that

$y=ut+\dfrac{1}{2}at^2$

$h+d=\dfrac{1}{2}gT^2$

$T=\sqrt{\dfrac{2(h+d)}{g}}$

$T=\sqrt{\dfrac{2(h+\dfrac{v^2}{2g})}{g}}$

So the total time = T+ t

$Total\ time\ is =\dfrac{v}{g}+\sqrt{\dfrac{2(h+\dfrac{v^2}{2g})}{g}}$