Question

A ball is thrown vertically upward from a height of $40\text{m}$ and hits the ground with a speed that is three times it's initial speed. What is the time taken (in sec.) for the fall?

• A. $2\text{sec}$
• B. $3\text{sec}$
• C. $4\text{sec}$
• D. $8\text{sec}$

Answer 1

The correct option is C $4\text{sec}$

Given, height from which ball is thrown $h=40\text{m}$
We have from the equation
$S=ut+\frac{1}{2}a{t}^2$
$\Rightarrow -40=ut-\frac{1}{2}\times 10\times t^2$
$\Rightarrow -40=ut-5t^2$ ...(1)

Now from $3^{rd}$ equation of motion, we have
$v^2=u^2+2as$
$\Rightarrow (3u{)}^2=u^2+2(-10\left)\right(-40)$
$\Rightarrow 9u^2=u^2+800$
$\Rightarrow u=10\text{m/s}$

Now, put value of $u$ in equation (1)
$\Rightarrow -40=10t-5t^2$
$\Rightarrow t^2-2t-8=0$
$\Rightarrow t^2-4t+2t-8=0$
$\Rightarrow t(t-4)+2(t-4)=0$
$\Rightarrow (t+2)(t-4)=0$
$\Rightarrow t=-2,4$
Hence, the time taken for the fall is $4\text{sec}$