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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics2-D & 3DMedium2 minQB

PhysicsMediumnumerical

A projectile with fixed muzzle speed falls 30 m short of a mark at elevation 15° and overshoots it by 70 m at elevation 45°. The proper elevation to hit the mark is about:

Answer:

20.30

Solution:

$sin 2 θ = \frac{70 \times s i n 3 0 ^{\circ} + 30 \times s i n 9 0 ^{\circ}}{100} = \frac{35 + 30}{100} = 0.65$ ,
so
$θ = \frac{1}{2} sin^{- 1} (0.65) = 20. 3^{\circ}$ .

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: QB

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