Question

# A healthy young man stand at a distance of $7m$ from a $11.8m$height building sees a kid slipping from the top floor. With what speed (assume uniform) should he run to catch the kid at the arm s height at $1.8m$

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Solution

## Step 1. Given dataA healthy young man stand at a distance of $7m$. The height of the building is $11.8m$.The kid should caught over $1.8m$ from the ground level. Initial velocity is $0m/s$Step 2. Formula usedSpeed =$\frac{s}{t}$where $s$ is the distance and $t$ is the timeEquation of the motion $s=ut+\frac{1}{2}g{t}^{2}$where $s$is the distance , $u$ the initial velocity $g$is the gravitational constant.(take $g=9.8m/{s}^{2}$) and $t$ is the time. Step 3. Find the distance traveled by the kidThe distance covered by the kid before he was caught is $=11.8-1.8=10m$Step 4. Find the time taken to fallLet the time taken to fall is $t$$s=10m\phantom{\rule{0ex}{0ex}}g=9.8m/{s}^{2}\phantom{\rule{0ex}{0ex}}u=0m/s$The equation of the motion is $s=ut+\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}⇒10=0+\frac{1}{2}×9.8×{t}^{2}\phantom{\rule{0ex}{0ex}}⇒{t}^{2}=2.04\phantom{\rule{0ex}{0ex}}⇒t=1.42s$Step 5. Find the speed at which the man ran The speed at which the man ran in order to catch the kid is Speed$=\frac{s}{t}$ $=\frac{7}{1.42}\phantom{\rule{0ex}{0ex}}=4.9m/s$Hence, the man ran at a speed of $4.9m/s$.

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