Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics2-D & 3DEasy2 minPYQ_2020

PhysicsEasynumerical

A particle is projected from the earth's surface with a velocity of $50 m s^{- 1}$ at an angle $θ$ with the horizontal. After $2 s$ it just clears a wall $5 m$ high. What is the value of $55 \sin θ$ ? ( $g = 10 m s^{- 2}$ )

Answer:

13.75

Solution:

Using equation of motion in y-dir $s = u t + \frac{1}{2} a t^{2}$

Height after $2 s$ ,

$h = 5 = (50 sin θ) t - \frac{1}{2} g {(t)}^{2}$

$5 = (50 sin θ) (2) - \frac{1}{2} \cdot 10 \cdot {(2)}^{2}$

$\frac{25}{100} = sin θ$

$sin θ = \frac{1}{4}$

$55 sinθ = \frac{55}{4} = 13.75$

Subject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: PYQ_2020

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