a projectile at an angle of 60 degree. if its time of flight 16 sec . calculate its horizontal range
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Explanation:
At any time t, the 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
Vx = VCosθ
Vy = VSinθ – gt
At maximum height, Vy = 0 = VSinθ – gt
So at maximum height, t = (VSinθ)/g
Maximum height occurs at ½ total flight time
This occurs when t = 8 = VSin60/g so V = 8(9.81)/0.866) = 90.62m/s
The range R of a projectile launched at an angle θ with a velocity V is:
R = V^2 Sin2θ / g
So the Range is (90.62^2)Sin120/9.81 = 724.95m
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