MHT CET · Physics · Motion In Two Dimensions
A projectile thrown from the ground has initial speed ' \(u\) ' and its direction makes an angle ' \(q\) ' with the horizontal. If at maximum height from ground, the speed of projectile is half its initial speed of projection, then the maximum height reached by the projectile is
\(\text {[g }=\text { acceleration due to gravity, } \sin 30^{\circ}=\cos 60^{\circ}\) \(=0.5,\) \(\cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2]\)
- A \(\frac{2 \mathrm{u}^2}{\mathrm{~g}}\)
- B \(\frac{3 u^2}{8 g}\)
- C \(\frac{\mathrm{u}^2}{\mathrm{~g}}\)
- D \(\frac{\mathrm{u}^4}{2 \mathrm{~g}}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 u^2}{8 g}\)
Step-by-step Solution
Detailed explanation
Given at maximum height
\(\begin{aligned} & \mathrm{u} \cos \theta=\frac{1}{2} \mathrm{u} \Rightarrow \cos \theta=\frac{1}{2} \\ & \therefore \theta=60^{\circ}\end{aligned}\)
We know, \(\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\mathrm{u}^2 \sin ^2 60^{\circ}}{2 \mathrm{~g}}=\frac{3 \mathrm{u}^2}{8 \mathrm{~g}}\)
\(\begin{aligned} & \mathrm{u} \cos \theta=\frac{1}{2} \mathrm{u} \Rightarrow \cos \theta=\frac{1}{2} \\ & \therefore \theta=60^{\circ}\end{aligned}\)
We know, \(\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\mathrm{u}^2 \sin ^2 60^{\circ}}{2 \mathrm{~g}}=\frac{3 \mathrm{u}^2}{8 \mathrm{~g}}\)
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