Question

A ball is thrown upwards. Its height varies with time as shown in figure. If the acceleration due to gravity is $7.5m/s^2$, then the height $h$ is

• A. 10 m
• B. 15 m
• C. 20 m
• D. 25 m

Answer 1

Answer: (B)

The height $h$ is covered in time interval $t=1$ s to $t=$
$2s$ or $t=5s$ to $t=6s$
Let $u$ be the initial velocity given to the ball when projected vertically upwards. Then
$h=S_2-S_1=S_5-S_6$
$S_2-S_1=\left(u\times 2-\frac{1}{2}\times 7.5\times 2^2\right)-\left(u\times 1-\frac{1}{2}\times 7.5\times 1^2\right)$
$=u-\frac{3}{2}\times 7.5$
$S_5-S_6=\left(u\times 5-\frac{1}{2}\times 7.5\times 5^2\right)-\left(u\times 6-\frac{1}{2}\times 7.5\times 6^2\right)$
$=-u+\frac{11}{2}\times 7.5$
$\therefore u-\frac{3}{2}\times 7.5=-u+\frac{11}{2}\times 7.5$
or $2u=\frac{14}{2}\times 7.5=7\times 7.5$ or
$u=7\times \frac{7.5}{2}=26.25m/s$
$\therefore h=26.25-\frac{3}{2}\times 7.5$
$=26.25-11.25=15.0m$