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## Homework Statement

A cannon located 60m from the base of a vertical 25 m cliff shoots a shell at 43 degrees above the horizontal.

What is the minimum muzzle velocity be for the shell to clear the top of the cliff ?

The ground at the top of the cliff is level, with constant elevation of 25 m above the cannon, How far does the shell land past the cliff ?

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A man stands on the roof of a 15 m tall building a throws a rock with a velocity of magnitude 30 m/s at an angle of 33 degrees above the horizontal.

What is the maximum height above the roof reached by the rock .?

The magnitude of the velocity of the rock before it strikes the ground ?

The horizontal distance from the base of the building to the point when the rock strikes the ground .?

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A ball is thrown upward with an initial spped of 20 m/s from the edge of a 45 m high cliff. At the istant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6 m/s .

At what angle above the horizontal should the ball be thrown so the runner will catch it just before it hits the ground and how far does the women wrun before she catches the ball .?

## Homework Equations

Vy=vsin

Vx=vcos

dx= vx * t

dy = Vyt - .5at^2

## The Attempt at a Solution

First problem I do not know where to start.

Second problem I tried and got the answers wrong to all 3. I first solved the velocities for the x and y directions. Then using Vy found out how long it took to hit the ground then use half of that to find the maximum height. But i got that wrong and the rest followed.

Third problem is used 6 m/s = Vx and then figured out the angle using Vx=Vcos then figured out Vy then I was stuck.