A gun and a target are at the same horizontal level separated by a distance of \(600 \mathrm{~m}\). The…
A gun and a target are at the same horizontal level separated by a distance of \(600 \mathrm{~m}\). The bullet is fired from the gun with a velocity of \(500 \mathrm{~ms}^{-1}\). In order to hit the target, the gun should be aimed to a height \(h\) above the target. The value of \(h\) is (Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )
\(2.4 \mathrm{~m}\)
\(3.6 \mathrm{~m}\)
\(7.2 \mathrm{~m}\)
\(10.8 \mathrm{~m}\)
Solution
Given,
distance between gun and target \(=600 \mathrm{~m}\) velocity of bullet \(=500 \mathrm{~ms}^{-1}\)
Now, distance \(=\) velocity of bullet \(\times\) time
$\begin{aligned}
600 &= 500 \times t \\
t &= \frac{600}{500} = \frac{6}{5} = 1.2 \, \text{sec}
\end{aligned}$
From the second equation of motion,
$ h = u t + \frac{1}{2} g t^{2} $
Putting the given values, we get
$\begin{aligned}
h & = 0 \times (1.2) + \frac{1}{2} \times (-10) \times (1.2)^{2} \\
h & = \frac{1}{2} \times (-10) \times (1.2 \times 1.2) \\
h & = -7.2 \, \text{m} \\
|h| & = 7.2 \, \text{m}
\end{aligned}$