Question

A ball is projected with a velocity 30 m/s making an angle 60 with the horizontal. The velocity at the highest point is​

Answer 1

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

In this case, the Vy = 0 at maximum height, so at that point,

V(max height) = Vx = VCosθ = 30Cos60° = 30(0.5) = 15m/s

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