Question

a boy drop a ball from the top of tower of height 19.6m. calculate the velocity of ball just before it touches the ground

Answer 1

Using formula: v^2-u^2=2as
Initial velocity(u)=0
Distance(s)=19.6m
Acceleration(a) = 9.6 (due to gravity)
Therefore substituting the values
v^2= 2(9.8)(19.6)
v^2=(19.6)(19.6)
v^2=(19.6)^2
thus, v= 19.6 m/s

Answer 2

_/\_Hello mate__here is your answer--

_____________________________

u = 0 m/s

v = ?

s = Height of the stone = 19.6 m

g = 9.8 ms−2

According to the equation of motion under gravity

v^2 − u^2 = 2gs

⇒ v^2 − 0^2 = 2 × 9.8 × 19.6

⇒ v^2 = 2 × 9.8 × 19.6 = (19.6)^2

⇒ v = 19.6 ms−1

Hence, the velocity of the stone just before touching the ground is 19.6 ms^−1.

I hope, this will help you.☺

Thank you______❤

__________________________❤