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Question

A projectile is thrown at an angle θ from the horizontal with velocity 'u' under the gravitation field of the earth. Derive expressions for its:
a)Time of its flight
b)Height
c)Horizontal Range

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Solution


Step 1: Time of flight calculation [Refer Fig.]
For calculation of time of flight, we always apply equations of motion in vertical direction.

Since vertical acceleration g is constant, therefore applying equation of motion in y direction

Motion from O to B:
Sy= Total displacement in y=0; uy=usinθ; ay=g
Sy=uyt+12ayt2
0=usinθ t12gt2

t=2usinθg

Step 2: Horizontal range in x direction

ux=ucosθ;ax=0;sx=R

Since acceleration is zero in x direction

So, Sx=uxt

R=ucosθ×2usinθg

R=u2sin2θg

Step 3: Maximum height
From Figure, at maximum height only horizontal component of velocity is present, vertical component becomes zero,

So, Applying eqn of motion in y direction between OA
uy=usinθ;vy=0,ay=g, sy=H

v2yu2y=2aysy

02(usinθ)2=2gH

H=u2sin2θ2g

2111223_1190325_ans_7cd13d7e194c4176bcbab2677ce1729a.png

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