Question

A ball is thrown vertically downwards from a tower of height $60\text{m}$ with a speed of $20\text{m/s}$. The $y\text{(m)}-t$ curve of the ball is
[Take $g=10{\text{m/s}}^2$, consider point of projection as origin and upward displacement as positive]

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Given, height of the building $h=60\text{m}$, initial speed of the ball $u=20\text{m/s}$
The velocity of the ball is given as
$v=u+at\Rightarrow v=-20-10t$
Now, we will find the time of flight of the ball using the equation $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow -60=-20t-\frac{1}{2}\times 10\times t^2$
On solving the equation we get $t=2\text{sec}$
So, the ball will take $2\text{sec}$ to reach the ground
Hence, the striking velocity of the ball will be
$v=-20-10\times 2=-40\text{m/s}$
Thus, the situation can be shown as

The $v-t$ graph will be

So, the $y\text{(m)}-t$ curve will be