Question

A stone is released from the top of a tower of height 78.4 m. Find the time taken by the stone to reach the ground.

Answer 1

Step 1: Given

Initial velocity of the ball, $\left(u\right)=0$

Height $\left(h\right)=78.4$

Acceleration due to gravity$\left(g\right)=9.81$

Let t is the time taken by stone to reach the ground.

Step 2: Formula used
From second equation of motion,
$h=ut+\frac{1}{2}g{t}^2$

Step 3: Finding the time

$h=ut+\frac{1}{2}g{t}^2$

Putting the values in the equation,
$\Rightarrow h=\frac{1}{2}g{t}^2$
$\Rightarrow 78.4=\frac{1}{2}\times 10\times t^2$
$\Rightarrow t=3.95s$
Hence, the time taken by the stone to reach on the ground is $3.95s$.