Question

A ball is thrown upwards. Its height varies with time as shown in figure. If the acceleration due to gravity is $7.5 \,m / s ^{2}$, 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 $t=$
$2 s$ or $t=5 s$ to $t=6 s$
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 $2 u=\frac{14}{2} \times 7.5=7 \times 7.5$ or
$u=7 \times \frac{7.5}{2}=26.25 m / s$
$\therefore h=26.25-\frac{3}{2} \times 7.5$
$=26.25-11.25=15.0\, m$