A ball is thrown vertically up from a tower of height $60\text{m}$ with a speed of $20\text{m/s}$. The $v-t$ curve of the ball is
[Take upward positive and $g=10{\text{m/s}}^2$]

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
$v=u+at\Rightarrow v=20-10t$
So, at $t=2\text{sec}$, the ball will be at maximum height and velocity will be zero.
Now, we will find the time of flight of the ball using the equation $s=ut+\frac{1}{2}a{t}^2$ and $T=\frac{2u}{g}$
So, time taken to cover building height is given by
$\Rightarrow -60=-20t-\frac{1}{2}\times 10\times t^2$
On solving the equation we get $t=2\text{sec}$
Also, time taken by ball to come at the same level from where it was thrown is given by
$T=\frac{2u}{g}=\frac{2\times 20}{10}=4\text{sec}$
So, the ball will take $2+4=6\text{sec}$ to reach the ground
Hence, the striking velocity of the ball will be
$v=20-10\times 6=-40\text{m/s}$
Thus, the situation can be shown as

The $v-t$ graph would be