Question

A ball is thrown vertically upward from the top of a tower with an initial velocity of 19.6 metre per second. the ball reaches the ground after 5 seconds , calculate the height of the tower and the velocity of ball on reaching the ground
$take \: g \: = 9.8mps2$
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Answer 1

Answer: h=24.5m

           final velocity,v= 21.91m/s

Explanation: Let the height of the tower = h

 Given,

                    initial velocity,u=19.6m/s

time ,t=5s

we know

          s=ut + 1/2gt²

        ⇒ -h = 19.6×5 + 1/2×(-9.8)×5²      

        (s=displacement=final posn - initial position=0 - h = -h)

(here g is taken negative as the motion is upward)

after solving we get

     h=24.5m

2nd part

v² - u²= 2gs

v² = 2×(-9.8)(-h)

v=21.91m/s