Question

A ball is thrown vertically upward from top of a tower with a velocity 10 M per second if the ball falls on a ground after 5 second the height of the tower will be

Answer 1

u=10m/s
time=5s
take g =10m/s ²
s=ut -½gt²
50- ½*10*25
50-125=75

Answer 2

Answer:

The height of the tower would be 72.5 m

Explanation:

A ball is thrown vertically upward from top of a tower with a velocity 10 m/s

If the ball falls on a ground after 5 second.

Using motion equation:

$S=ut-\dfrac{1}{2}gt^2+h_0$

where,

u → initial velocity ( u=10 m/s )

t → Time to reach ground ( t=5 s )

h₀ → height of tower ( h )

S →  vertical distance from ground ( S=0 )

g → acceleration due to gravity ( g=9.8 m/s² )

By substituting the these value into formula and we get,

$0=10\cdot 5-\dfrac{1}{2}\cdot 9.8\cdot 5^2+h$

$0=50-122.5+h$

$h=72.5\ m$

Hence, The height of the tower would be 72.5 m