Question

A ball is thrown vertically upwards with an initial velocity such that it can reach a maximum height of 15 m, if at the same instance a stone is dropped from the same height of 15 m, find the ratio of distances travelled by them when they cross each other?

• A. $4:3$
• B. $2:3$
• C. $1:2$
• D. $3:1$

Answer 1

The correct option is D $3:1$

Let initial velocity of ball is $u$.

At maximum height , velocity $v=0$,

Using $v^2=u^2+2as$ $\Rightarrow 0=u^2-2g\times 15$ $\Rightarrow u=\sqrt{2\times 10\times 15}$,

Let $h_1$ distance travelled by ball and $h_2$ distance travelled by stone.

Let at time $t$ they crossed each other.

Using $s=ut+\frac{1}{2}a{t}^2$ $\Rightarrow h_1=ut-\frac{1}{2}g{t}^2$ $\Rightarrow h_1=\sqrt{2\times 10\times 15}t-\frac{1}{2}g{t}^2$ ....1

$\Rightarrow -h_2=0\times t-\frac{1}{2}g{t}^2$ $\Rightarrow h_2=\frac{1}{2}g{t}^2$ ....2

Adding equations 1 and 2, $\Rightarrow h_1+h_2=15=\sqrt{2\times 10\times 15}t-\frac{1}{2}g{t}^2+\frac{1}{2}g{t}^2$

$\Rightarrow \sqrt{2\times 10\times 15}t=15$ .....3

$\Rightarrow t^2=\frac{15}{20}$ $\Rightarrow \frac{1}{2}g{t}^2=\frac{15}{4}$ ...4

Using equations 3 and 4 , $\Rightarrow h_1=\frac{45}{4}$ and $h_2=\frac{15}{4}$

$\therefore \frac{h_1}{h_2}=\frac{\frac{45}{4}}{\frac{15}{4}}=\frac{3}{1}$ or $3:1$