Physics, asked by jyotiadv9415, 1 month ago

a projectile at an angle of 60 degree. if its time of flight 16 sec . calculate its horizontal range​

Answers

Answered by ak8842340
1

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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