Question

A body travel from tower and reach ground and take 4sec to reach ground find height of tower

Answer 1

GIVEN:-

• $\rm{Initial\:Velocity = 0m/s}$

• $\rm{Time\:Taken = 4s.}$

• $\rm{Acceleration\:due\:to\:gravity = +10m/s^{-2}}$

TO FIND:-

• Height of the tower.

FORMULAE USED:-

• ${\huge{\boxed{\large{\rm{S = ut + \dfrac{1}{2}\times{a}\times{(t)^2}}}}}}$.

Now,

$\implies\rm{h = ut + \dfrac{1}{2}\times{g}\times{(t)^2}}$

$\implies\rm{h = 0(t) + \dfrac{1}{\cancel{2}}\times{\cancel{10}}\times{(4)^2}}$

$\implies\rm{ h = 0 + 5+ 16}$

$\implies\rm{ h = 21m}$

Hence, The Height of tower is 21m.

EXTRA INFORMATION:-

• When object is at heigest position then Final Velocity is zero.

• When object is Projected to free fall then Initial Velocity is zero.

Answer 2

Given ,

Initial velocity (u) = 0 m/s

Time (t) = 4 sec

Acceleration due to gravity (a) = 10 m/s

We know that , the Newton's second equation of motion is given by

$\boxed{ \sf{s = ut + \frac{1}{2} a {(t)}^{2} }}$

Thus ,

s = 0(4) + 1/2 × 10 × (4)²

s = 0 + 5 + 16

s = 21 m

Therefore ,

• The height of the tower is 21 m