Question
From a 20 m high tower, a bullet is fired from a gun in the horizontal direct. If $g=10 ms^{-2}$, and initial velocity of bullet is $400 ms^{-1}$ then at what distance will be bullet fall on the earth?
✓
Answer
Let the time taken by bullet to fall in horizontal direction $=t sec$
Here,
$\begin{array}{l}
u=0 \\
h=\frac{1}{2} g t^2
\end{array}$
$\begin{aligned} 20 & =\frac{1}{2} \times 10 \times t^2 \\ t^2 & =4, \therefore t=2 sec \end{aligned}$
Distance of bullet falling on earth from tower :
$\begin{aligned}
BC & =u \times t \\
& =400 \times 2 \\
& =800 m .
\end{aligned}$
Here,
$\begin{array}{l}
u=0 \\
h=\frac{1}{2} g t^2
\end{array}$
$\begin{aligned} 20 & =\frac{1}{2} \times 10 \times t^2 \\ t^2 & =4, \therefore t=2 sec \end{aligned}$
Distance of bullet falling on earth from tower :
$\begin{aligned}
BC & =u \times t \\
& =400 \times 2 \\
& =800 m .
\end{aligned}$
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