Projectiles Question -- Tiger Woods drives a golf ball on the Moon

[rr96]

Homework Statement

As part of a NASA experiment, golfer Tiger Woods drives a golf ball on the moon, where
g = 1.60 m/s2. He ‘launches’ a golf ball with a speed of 285 km/h, at an angle of 42o with the horizontal. What horizontal distance will his drive travel before landing back on the surface of the moon. Ignore the curvature of the moon.

Homework Equations

d = (Vi)(t) + 1/2(a)t2

Vf =Vi + at

(Vf)2 = (Vi)2 + 2ad

The Attempt at a Solution

Initial horizontal velocity:

285 x cos42
= 211.8 km/h

Initial Vertical Velocity

285 x sin42
= 190.7 km/h

Finding time using vertical components

d = (Vi)(t) + 1/2(a)t2

0 = 190.7t - 4.9t2

t = 38.9 s

Using time to find distance

d = (Vi)(t) + 1/2(a)t2

d = 211.8 x 38.9 + 1/2(0)t2

d = 8239 m

 

[haruspex]

0 = 190.7t - 4.9t2

4.9?

 

[rr96]

Sorry, I skipped a step. 1/2 x 9.8 = 4.9

 

[Nathanael]

Sorry, I skipped a step. 1/2 x 9.8 = 4.9

Haruspex is a smart guy, I'm sure he could see the step you skipped.

His next question would be:

"9.8?"

 

[Nathanael]

Besides that, though, there's one more small problem. Your units are inconsistent.

(You need to convert km/hr to meters/second)

 

[rr96]

Thanks! I completely missed that. My final answer is 635 m

 

[Nathanael]

Thanks! I completely missed that. My final answer is 635 m

My answer disagrees.

How long will the ball be in the air? (What was your calculation for this?)

 

[rr96]

10.8 s ?

 

[rr96]

d = (Vi)(t) + 1/2(a)t2

0 = 52.97t - 4.9t2

t = 10.8 s

 

[Nathanael]

Have you forgotten where this is taking place? :) Remember, we're not on Earth.

 

[rr96]

d = (Vi)(t) + 1/2(a)t2

0 = 52.97t - 0.8t2

t = 66.21 s

I had forgotten! Thank you so much!

 

[rr96]

Final answer is 3.9 km

 

[Nathanael]

There you go, that should be the correct answer.

(Do you have something that says what the correct answer is?)

 

[rr96]

Yup! That's what it says the answer is.

 

[Dumbledore211]

You have missed out on crucial piece of information which the value of is 1.60m/s^2 as the event is taking place on the moon

 

[hr14]

Final answer is 3.9 km

Can you explain how you got this? I followed the replies and tried doing it myself but don't understand at all how you're getting to 3.9

 

[PeroK]

Can you explain how you got this? I followed the replies and tried doing it myself but don't understand at all how you're getting to 3.9

What do you get and why?

 

[hr14]

What do you get and why?

This is what I did...

vx: 285 x cos42
= 211.8 km/h = 58.83 m/s

vy: 285 x sin42
= 190.7 km/h = 52.97 m/s

d = (Vi)(t) + 1/2(a)t2
0 = 52.97t - 0.8t2
t = 66.2125s

d = (Vi)(t) + 1/2(a)t2
d = 58.83 x 66.2125 + 1/2(1.6)(66.2125)2
d = 388 m

388m = 0.388km

 

[haruspex]

d = (Vi)(t) + 1/2(a)t2
d = 58.83 x 66.2125 + 1/2(1.6)(66.2125)2

Gravity acts diagonally on the moon?!
But that expression gives 7402m.
I've played around with variants of the above expression for d (ignoring the quadratic term, making it negative) and none lead to 388.

 

[hr14]

Gravity acts diagonally on the moon?!
But that expression gives 7402m.
I've played around with variants of the above expression for d (ignoring the quadratic term, making it negative) and none lead to 388.

I tried again and did get 7402m but converting that to km doesn't give 3.9km (which is the correct answer).
I meant to put a negative sign instead of positive. But doing (-) instead of (+) should give 388

 

[haruspex]

I tried again and did get 7402m but converting that to km doesn't give 3.9km (which is the correct answer).
I meant to put a negative sign instead of positive. But doing (-) instead of (+) should give 388

You seem to have overlooked my rhetorical question:

Gravity acts diagonally on the moon?!

 

[hr14]

You seem to have overlooked my rhetorical question:

No, it doesn't. It acts vertically

 

[haruspex]

No, it doesn't. It acts vertically

So why do you have a gravity term in the horizontal displacement equation?

d = (Vi)(t) + 1/2(a)t2
d = 58.83 x 66.2125 + 1/2(1.6)(66.2125)2

 

[hr14]

So why do you have a gravity term in the horizontal displacement equation?

I figured it out. Thanks