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NEET PHYSICSEasy

Two bodies are projected with the same velocity. If one is projected at an angle of 30∘30^{\circ} and the other at an angle of 60∘60^{\circ} to the horizontal, the ratio of the maximum heights reached is:

A

3 : 1

B

1 : 3

C

1 : 2

D

2 : 1

Step-by-Step Solution

  1. Formula for Maximum Height: The maximum height HH reached by a projectile is given by the equation: H=v02sin⁡2θ2gH = \frac{v_0^2 \sin^2 \theta}{2g} .
  2. Comparison: Since the initial velocity v0v_0 and acceleration due to gravity gg are the same for both bodies, the height is directly proportional to sin⁡2θ\sin^2 \theta. H∝sin⁡2θH \propto \sin^2 \theta
  3. Calculation: For the first body (θ1=30∘\theta _1 = 30^{\circ}): H1∝sin⁡230∘=(1/2)2=1/4H_1 \propto \sin^2 30^{\circ} = (1/2)^2 = 1/4. For the second body (θ2=60∘\theta _2 = 60^{\circ}): H2∝sin⁡260∘=(3/2)2=3/4H_2 \propto \sin^2 60^{\circ} = (\sqrt{3}/2)^2 = 3/4.
  4. Ratio: H1H2=1/43/4=13\frac{H_1}{H_2} = \frac{1/4}{3/4} = \frac{1}{3}. The ratio is 1 : 3.
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