Stone thrown off a cliff: height and angle

[moenste]

Homework Statement

A stone thrown horizontally from the top of a vertical cliff with velocity 15 m s^-1 is observed to strike the (horizontal) ground at a distance of 45 m from the base of the cliff. What is (a) the height of the cliff, (b) the angle the path of the stone makes with the ground at the moment of impact?

Answers: (a) 45 m, (b) 63.4°

Homework Equations

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The Attempt at a Solution

I used s = v*t to find t: 45 m = 15 m s^-1 * t -> t = 3 s

Then I used s = 1/2 (u + v)*t -> s = 1/2 (15 m s^-1 + 15 m s^-1)*3 s = 45 m. I guess velocity when t = 0 (u) should be also 15 m s^-1, equal to v = 15 m s^-1. But I am not sure about the solution.

For the angle if height and horizontal distance both are 45 meters then tan angle = 45/45 -> angle = 45. But it's not the book answer.

Any help please?

 

[SteamKing]

Homework Statement

A stone thrown horizontally from the top of a vertical cliff with velocity 15 m s^-1 is observed to strike the (horizontal) ground at a distance of 45 m from the base of the cliff. What is (a) the height of the cliff, (b) the angle the path of the stone makes with the ground at the moment of impact?

Answers: (a) 45 m, (b) 63.4°

Homework Equations

-

The Attempt at a Solution

I used s = v*t to find t: 45 m = 15 m s^-1 * t -> t = 3 s

Then I used s = 1/2 (u + v)*t -> s = 1/2 (15 m s^-1 + 15 m s^-1)*3 s = 45 m. I guess velocity when t = 0 (u) should be also 15 m s^-1, equal to v = 15 m s^-1. But I am not sure about the solution.

For the angle if height and horizontal distance both are 45 meters then tan angle = 45/45 -> angle = 45. But it's not the book answer.

Any help please?

You're not thinking too clearly here about what happens to the stone after it is released.

Does the stone keep flying away from the cliff horizontally? If not, why not?

If you have a velocity of say 10 m/s horizontally and a velocity of 5 m/s vertically, is the combined velocity (10 m/s + 5 m/s) = 15 m/s ?

 

[PeroK]

Homework Statement

A stone thrown horizontally from the top of a vertical cliff with velocity 15 m s^-1 is observed to strike the (horizontal) ground at a distance of 45 m from the base of the cliff. What is (a) the height of the cliff, (b) the angle the path of the stone makes with the ground at the moment of impact?

Answers: (a) 45 m, (b) 63.4°

Homework Equations

-

The Attempt at a Solution

I used s = v*t to find t: 45 m = 15 m s^-1 * t -> t = 3 s

That's correct.

Then I used s = 1/2 (u + v)*t -> s = 1/2 (15 m s^-1 + 15 m s^-1)*3 s = 45 m. I guess velocity when t = 0 (u) should be also 15 m s^-1, equal to v = 15 m s^-1. But I am not sure about the solution.

This seems muddled to me. You know the stone is falling for 3 seconds, so you need to find an equation to calculate the vertical distance fallen.

For the angle if height and horizontal distance both are 45 meters then tan angle = 45/45 -> angle = 45. But it's not the book answer.

Any help please?

To find the angle, you need to find the velocity (not speed) of the stone when it hits the ground. You are imagining that the stone travels in a straight line at $45°$, which it doesn't: it travels in a parabolic curve, where the angle is changing all the time.

 

[moenste]

Does the stone keep flying away from the cliff horizontally? If not, why not?

If you have a velocity of say 10 m/s horizontally and a velocity of 5 m/s vertically, is the combined velocity (10 m/s + 5 m/s) = 15 m/s ?

As PeroK said the stone travels in a parabolic curve. I think it's due to the gravity acting on the stone.

Shouldn't it be 10² + 5² = 125 -> Sq root 125 = 11.18?

This seems muddled to me. You know the stone is falling for 3 seconds, so you need to find an equation to calculate the vertical distance fallen.

To find the angle, you need to find the velocity (not speed) of the stone when it hits the ground. You are imagining that the stone travels in a straight line at $45°$, which it doesn't: it travels in a parabolic curve, where the angle is changing all the time.

s = ut + 1/2 * a * t²
s = 0 * 3 s + 1/2 * 10 m s ^-1 * 3² s
s = 45 m
?

 

[SteamKing]

As PeroK said the stone travels in a parabolic curve. I think it's due to the gravity acting on the stone.

Why, yes, gravity may have something to do with the path the stone takes after it is thrown off a cliff.

Shouldn't it be 10² + 5² = 125 -> Sq root 125 = 11.18?

Yes, this is how one would calculate the resultant velocity given the components of the velocity. Remember, velocity has both magnitude and direction.

s = ut + 1/2 * a * t²
s = 0 * 3 s + 1/2 * 10 m s ^-1 * 3² s
s = 45 m
?

What does s represent here, in terms of the path the rock takes after it is thrown off the cliff?

Remember, when the rock is thrown, it is traveling horizontally away from the edge of the cliff while it is dropping vertically.

 

[moenste]

Why, yes, gravity may have something to do with the path the stone takes after it is thrown off a cliff.

Well the stone has a velocity of 15 m s^-1 and it reaches the top using it, after that he loses height and goes downwards with the 10 m s^-1 gravity.

Yes, this is how one would calculate the resultant velocity given the components of the velocity. Remember, velocity has both magnitude and direction.

11.18 and 26.56°.

What does s represent here, in terms of the path the rock takes after it is thrown off the cliff?

Remember, when the rock is thrown, it is traveling horizontally away from the edge of the cliff while it is dropping vertically.

s is height of the cliff.