What Is the Final Velocity of a Projectile on an Elevated Green?

[Mr Davis 97]

Homework Statement

A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40.0° above the horizontal . It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the peed of the ball just before it lands.

Homework Equations

$v_x = v_{0x} = \left \| v_0 \right \|\cos\theta$
$v_y^2 = v_{0y}^2 + 2ay$
$v = \sqrt{v_x^2 + v_y^2}$

The Attempt at a Solution

We begin with $v = \sqrt{v_x^2 + v_y^2}$ since the magnitude of the final velocity is what we desire. Thus, we need to find $v_x$ and $v_y$. Therefore, we use $v_x = v_{0x} = \left \| v_0 \right \|\cos\theta$ since the velocity in the x-direction does not change from the initial (we're neglecting air resistance). Thus $v_x = (14 ~m/s)\cos40^{\circ} = 10.7~m/s$. Now,we find $v_y$, the final velocity in the y-direction. Since time is not metioned, we'll use $v_y^2 = v_{0y}^2 + 2ay$, specifically $v_y = -\sqrt{v_{0y}^2 + 2ay}$. Thus $v_y = -\sqrt{v_{0y}^2 + 2ay} = -\sqrt{(v\sin\theta)^2 + 2(-9.8~m/s^2)(3~m - 0~m)} = -4.71~m/s$. Therefore, the magnitude of the final velocity should be $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10.7~m/s)^2 + (-4.71~m/s)^2} = 11.7~m/s$. Is this the correct answer? If not what am I doing wrong and why?

 

[Doc Al]

$v_y^2 = v_{0y}^2 + 2ay$

You might want to write this as $v_y^2 = v_{0y}^2 + 2a\Delta y$

(You have the change in height being positive.)

Another way to double check your answer is to use energy principles.

 

[haruspex]

You might want to write this as $v_y^2 = v_{0y}^2 + 2a\Delta y$

Did you mean $v_{0y}^2 = v_{y}^2 + 2a\Delta y$?

 

[Doc Al]

Did you mean $v_{0y}^2 = v_{y}^2 + 2a\Delta y$?

No, I think I had it correct.

I believe he's using $v_{0y}$ to represent the y-component of the initial velocity.

 

[haruspex]

No, I think I had it correct.

I believe he's using $v_{0y}$ to represent the y-component of the initial velocity.

Sorry - my mistake.

 

[Mr Davis 97]

You might want to write this as $v_y^2 = v_{0y}^2 + 2a\Delta y$

(You have the change in height being positive.)

Another way to double check your answer is to use energy principles.

So is the final answer correct or incorrect? I did utilize a change in y since the final position is 3m and the initial is 0m; the change in that is just 3m. Is my approach correct or am I missing something?

 

[Doc Al]

I did utilize a change in y since the final position is 3m and the initial is 0m; the change in that is just 3m.

Oops... I just realized that I misread the problem. The green was 3m higher, not the tee. So your approach looks fine to me. I'll check the answer in a minute.

 

[Doc Al]

Yes, your answer is correct.

 

[Mr Davis 97]

Yes, your answer is correct.

Awesome! Thanks.