Solving a Ball Thrown from a Cliff: Angle and Distance

[yojayydee]

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 .?

 

[tiny-tim]

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 ?

Hi yojayydee!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[baba89]

I would first analyze the given information and use the principles of projectile motion to solve the problems presented.

1. For the first problem, the minimum muzzle velocity needed for the shell to clear the top of the cliff can be calculated using the formula v^2 = u^2 + 2as, where v is the final velocity (which in this case is 0 m/s as the shell reaches the top of the cliff), u is the initial velocity, a is the acceleration due to gravity (which is -9.8 m/s^2 for objects thrown upwards), and s is the displacement (which is 25 m in this case). Solving for u, we get u = 34.6 m/s as the minimum muzzle velocity needed for the shell to clear the top of the cliff.

To calculate the distance the shell lands past the cliff, we can use the formula x = ut + 1/2at^2, where x is the horizontal distance, u is the initial velocity (which in this case is 34.6 m/s), t is the time (which can be calculated using the given angle and initial velocity), and a is the acceleration due to gravity. Solving for x, we get x = 75.6 m as the distance the shell lands past the cliff.

2. For the second problem, the maximum height above the roof reached by the rock can be calculated using the formula y = u^2sin^2θ/2g, where y is the maximum height, u is the initial velocity (which is 30 m/s), θ is the angle of projection (which is 33 degrees), and g is the acceleration due to gravity. Solving for y, we get y = 15.9 m as the maximum height reached by the rock above the roof.

The magnitude of the velocity of the rock before it strikes the ground can be calculated using the formula v = u + at, where v is the final velocity (which in this case is 0 m/s as the rock strikes the ground), u is the initial velocity (which is 30 m/s), a is the acceleration due to gravity, and t is the time (which can be calculated using the given angle and initial velocity). Solving for v, we get v = 29.8 m/s as the magnitude of the velocity of the rock before it strikes the ground.

To calculate the horizontal distance from the base