Question

A ball is projected vertically downward with initial velocity 30 m/s from the top of the tower of height 80 m. Find the time taken by the ball to reach the ground.​

Answer 1

Given:

initial velocity 30 m/s

tower of height 80 m

 

To be found:

The time taken by the ball to reach the ground .

Formulas used

Kinematic equations for uniformly accelerated motion.

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and

$\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

Solution

By equation of motion

$\boxed{\bf S = ut +\frac{1}{2}a{t}^{2}}$

$\implies \bf 80= 30t +5t^2 \\ \\ \implies \bf 5(t^2+6t-16)=0\\ \\ \implies \bf t^2+6t-16=0 \\ \\ \implies \bf (t+8)(t-2)=0 \\ \\ \implies \bf \boxed{\bf{t}}\ can't\ be\ negative \\ \\ So\ \boxed{\boxed{\bf t =2}}$

Hence,

The time is taken by the ball to reach the ground = 2 sec