What is the speed of the ball when it hits the ground?

[JaZZyCooL]

Homework Statement

A ball is thrown from the edge of a 10.0 m high cliff with a speed of 20.0 m/s at an angle of 30 degrees below horizontal. What is the speed of the ball when its hits the ground below the cliff?

Options :
1) 26.3 m/s
2) 28.1 m/s
3) 31.4 m/s
4) 24.4 m/s
5) 29.5 m/s

I am new to the forum so bare with me I don't know how the forum works. I have attempted a solution for the problem. I would appreciate if you could just tell me the equation to work with.

Homework Equations

This is the equation I tried

v = sqrt(2*g*h)

The Attempt at a Solution

v = sqrt(2*9.8*10)
v = 14 m/s

The answer is not even in the options. Clearly I am doing something wrong actually way off. May be I am mixing wrong concept here.

I would appreciate explanation with equation.

Thank you.

 

[berkeman]

Homework Statement

A ball is thrown from the edge of a 10.0 m high cliff with a speed of 20.0 m/s at an angle of 30 degrees below horizontal. What is the speed of the ball when its hits the ground below the cliff?

Options :
1) 26.3 m/s
2) 28.1 m/s
3) 31.4 m/s
4) 24.4 m/s
5) 29.5 m/s

I am new to the forum so bare with me I don't know how the forum works. I have attempted a solution for the problem. I would appreciate if you could just tell me the equation to work with.

Homework Equations

This is the equation I tried

v = sqrt(2*g*h)

The Attempt at a Solution

v = sqrt(2*9.8*10)
v = 14 m/s

The answer is not even in the options. Clearly I am doing something wrong actually way off. May be I am mixing wrong concept here.

I would appreciate explanation with equation.

Thank you.

Welcome to the PF.

v = sqrt(2*g*h)

That would be the speed if the ball were dropped from the height h. But it was thrown, so there will be some sort of addition of speeds...

 

[Matejxx1]

A ball is thrown from the edge of a 10.0 m high cliff with a speed of 20.0 m/s at an angle of 30 degrees below horizontal. What is the speed of the ball when its hits the ground below the cliff?

v = sqrt(2*9.8*10)
v = 14 m/s

your ball also travels horizontally not just vertically try adding that to your equation

 

[JaZZyCooL]

So is it v=sqrt(2*9.8*10) +20
so V = 14 m/s + 20 m/s

v = 24 m/s?

 

[berkeman]

So is it v=sqrt(2*9.8*10) +20
so V = 14 m/s + 20 m/s

v = 24 m/s?

14+20 is not 24...

Besides, did you see the hint by Matejxx1? The velocity of the ball has horizontal and vertical components. Gravity affects the vertical component...

 

[JaZZyCooL]

v = sqrt (Vi^2 + 2*g*h)
v = sqrt( 20^2 + 2*9.8*10)

v = 24.4 m/s ?

 

[berkeman]

v = sqrt (Vi^2 + 2*g*h)
v = sqrt( 20^2 + 2*9.8*10)

v = 24.4 m/s ?

That looks good. You could have also solved it by finding the initial Vx and Vy components, then added in the extra Vy component due to the acceleration of gravity over the 10m drop, and then taken the SQRT of the squares of the x & y velocity components. Maybe try that method to be sure you get the same answer.

 

[phsicsgeek]

I calculated the initial x direction speed (17.4 m/s) and y direction (up & down) 10 m/s using simple trigonometry. The x direction speed does not change (assumes no air friction) and the y direction changes to 17.2 m/s calculated by adding the initial velocity (10 m/s) with the added velocity by dropping 10 m (t = 0.735 s drop). The square root of the sum of the squares yields the 24.4 m/s answer.

 

[jbriggs444]

I calculated the initial x direction speed (17.4 m/s) and y direction (up & down) 10 m/s using simple trigonometry. The x direction speed does not change (assumes no air friction) and the y direction changes to 17.2 m/s calculated by adding the initial velocity (10 m/s) with the added velocity by dropping 10 m (t = 0.735 s drop). The square root of the sum of the squares yields the 24.4 m/s answer.

You are responding to a thread that is seven years old.

The approach of separating the velocities is workable. But there is a shortcut. Conservation of energy. Mass is irrelevant. So pick a mass and call it $m$. You know starting speed. So you know starting kinetic energy. You know potential energy. So you know final kinetic energy. So you know final speed. No trig involved. [Work algebraicly. Derive formula. Evaluate formula. That way the mass $m$ drops out cleanly].

It'll still come out as the square root of a sum involving squares. You just do not have to separate out the squared initial horizontal velocity component from the squared initial vertical velocity component. Pythagoras already tells you the sum of those two squares.