Question

12. A projectile is projected at an angle of 60°, if its time of flight is 16
sec, calculate its horizontal range.​​

Answer 1

Answer:

A projectile is projected at an angle of 60°, if its time of flight is 16 seconds, what is its horizontal range?

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