Q123 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Physics
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, \(g=10 \mathrm{~ms}^{-2}\) )
- A\(2.4 \mathrm{~m}\)
- B\(3.6 \mathrm{~m}\)
- Correct answerC\(7.2 \mathrm{~m}\)
- D\(10.8 \mathrm{~m}\)
Solution
Given,
distance between gun and target \(=600 \mathrm{~m}\) velocity of bullet \(=500 \mathrm{~ms}^{-1}\)
Now, distance \(=\) velocity of bullet \(\times\) time
\(\begin{gathered}
600=500 \times t \\
t=\frac{600}{500}=\frac{6}{5}=1.2 \mathrm{sec}
\end{gathered}\)
From second equation of the motion,
\(h=u t+\frac{1}{2} g t^2\)
Putting the given values, we get
\(\begin{aligned}
h & =0 \times(1.2)+\frac{1}{2} \times(-10) \times(1.2)^2 \\
h & =\frac{1}{2} \times(-10)(1.2 \times 1.2) \\
h & =-7.2 \mathrm{~m} \\
|h| & =7.2 \mathrm{~m}
\end{aligned}\)
Hence, the correct answer is option: 'C'.