Question

Find an expression for the time of flight, maximum height and horizontal range of a projectile fired at an
angle with the horizontal. When is horizontal range maximum?

Answer 1

Answer:

What is the expression when a projectile fired from an angle theta with the vertical direction find the maximum height time of flight horizontal range velocity of projectile at any time t?

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My notes from “Projectiles 101”. These equations connect the variables of projectile motion, and may be useful to you.

At any time t, a projectile's horizontal and vertical displacement are, using V as initial velocity and θ as the angle relative to the horizontal:

x = VtCos θ

y = VtSinθ – ½gt^2

[If you really want to use an angle relative to the vertical, use (90° - θ) in the equations]

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

The velocity at any time t will be √(Vx^2 + Vy^2)

At maximum height, Vy = 0 = VSinθ – gt

So at maximum height, t = (VSinθ)/g (this is half the total flight time)

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

These equations neglect air resistance, of course. Otherwise, they become considerably more complicated!