Question

A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 41.2 m above the ground. The projectile strikes the ground with a speed of 1.1v0. Find v0. (Ignore any effects due to air resistance.)

Answer 1

Answer:

Explanation:

To do this, we need to find the angle ϕ

as it lands on the ground. Solving, we have

V0cos(θ)=Vfcos(ϕ)

V0cos(θ)=1.1V0cos(ϕ)

cos(30)=1.1cos(ϕ)

ϕ=38∘

Calculating for the final velocity, we have

Vfsin(ϕ)=√2aS

Vfsin(38)=√2×9.8×41.2

Vf=46.16ms

And now, we have the initial velocity V0

that is

1.1V0=Vf (V0 is Vnot)

V0=46.16/1.1

V0=41.96ms

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Answer 2

Only force acting on the body is gravitational force. Reduction in potential energy causes increase in kinetic energy so

Answer is 62.6m/s