derive expression for time of light of horizontal range of projectile
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Answer:At any time t, a projectile's horizontal and vertical displacement are:
x = VtCos θ where V is the initial velocity, θ is the launch angle
y = VtSinθ – ½gt^2
The velocities are the time derivatives of displacement:
Vx = VCosθ (note that Vx does not depend on t, so Vx is constant)
Vy = VSinθ – gt
At maximum height, Vy = 0 = VSinθ – gt
So at maximum height, t = (VSinθ)/g [total flight time = 2t]
The range R of a projectile launched at an angle θ with a velocity V is:
R = V^2 Sin2θ / g
The maximum height H is
H = V^2 Sin^2(θ) / 2g
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plz refer the picture for your answer.
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