Question

An aeroplane flying horizontally at a height of $980m$ with velocity $100m/s$ drops a food packet. A person on the ground is $100(\sqrt{2}-1)$ $m$ ahead horizontally from the dropping point. At what velocity approximately should he move so that he can catch the food packet.

• A. $50\sqrt{2}m{s}^{-1}$
• B. $\frac{50}{\sqrt{2}}m{s}^{-1}$
• C. $100m{s}^{-1}$
• D. $200m{s}^{-1}$

Answer 1

Answer: (C)

$100m{s}^{-1}$

For food packet
$u_h=100m/s$
$u_v=0$
$s_v=-980m$
$g=-10m/s^2$
$s=u_{v}t+\frac{1}{2}a{t}^2$
$(-980)=0+\frac{1}{2}(-10)t^2$
$\Rightarrow t=14s$

Distance travelled horizontally in $14$ s by food packet
$s_h=u_h\times t=100\times 14=1400m$

Person position is $100(\sqrt{2}-1)$ ahead of food packet at time of dropping:

$=41.42m$
distance person has to run$=1400-41.42$ $=1358.57m$ in time $=14s$

so his speed will be $=\frac{1358.57}{14}=97m/s$
$\sim 100m/s$