Solve Kinematics Q13: Golf Ball Height & Distance

[John A]

13) A golfer hits a ball at 100 m/s with an angle of elevation of 30 degrees towards a green. The acceleration due to gravity is 9.8 m/s^2 [down]. There is no resistance.

a) If the green is at the same height as the golfer then find how long the ball is in the air, and how far away the ball lands from where it was hit.

Given: Gravity: 9.8m/s^2, Vi= 100 m/s
What I've solved for: (x direction) di= 0m, df= 441.66 m, Vi= 86.6 m/s, Vf= 0m/s, a= 0 m/s^2, t=5.1s
(y-dir) di= 0m, df=127.5m, Vi=50 m/s, Vf= 0 m/s, a= -9.8 m/s^2, t= 10.2s)
I solved for the final x and y direction distances using trig laws.

What I'm having trouble with:
b) If the green is 10 m higher than the golfer then find out how long the ball is in the air, and how far away the ball lands. [You do not need to recalculate what you did in Part A]

My Answer (which was wrong):
Since initial and final velocity remain the same, the ball will be in the air for the same amount of time as Part A and should land only 10 m higher than that of Part A which is 137.5 m in the y-dir and still 441.66 m in the x-dir.

What am I doing Wrong?

 

[TomHart]

What if the landing spot was 10,000 feet lower. Would the hang time still be the same? And does that affect the x distance?

 

[TomHart]

John A, I apologize, but I should have looked at your work more carefully. I agree with your initial velocities that you found, but how did you come up with the x direction final velocity being equal to 0 m/s? Maybe that was my tee shot on hole #4 years ago when I hit that lady on her back side. But that would be more of an inelastic collision problem. But seriously, the x direction final velocity cannot be 0 m/s when the problem specifies "no resistance". And I agree with your time of t = 10.2 seconds for the y direction, but how can you get an answer of half that time for the x direction? Doesn't it have to be the same amount of time for x and y components that the ball is in the air?

 

[haruspex]

Since initial and final velocity remain the same,

Do they?

 

[John A]

John A, I apologize, but I should have looked at your work more carefully. I agree with your initial velocities that you found, but how did you come up with the x direction final velocity being equal to 0 m/s? Maybe that was my tee shot on hole #4 years ago when I hit that lady on her back side. But that would be more of an inelastic collision problem. But seriously, the x direction final velocity cannot be 0 m/s when the problem specifies "no resistance". And I agree with your time of t = 10.2 seconds for the y direction, but how can you get an answer of half that time for the x direction? Doesn't it have to be the same amount of time for x and y components that the ball is in the air?

The ball has to come to a stop, therefore final velocity in both x & y direction must equal 0 m/s

 

[Merlin3189]

The ball has to come to a stop, therefore final velocity in both x & y direction must equal 0 m/s

Aah! There's our issue.
All these equations of motion apply to the free flight of the ball through the air. As soon as it hits the ground, everything changes and other factors come into play. The question asks about the time in the air and where and when it hits the ground - nothing after that. Up to that point it is still moving, both horizontally and vertically. It will only stop after hitting the ground, probably bouncing and rolling, loosing energy to the ground. (I think the "no resistance" is probably meant to apply to air resistance. Otherwise, if there were truly no resistance of any sort, the ball would never stop!)

So your final zero velocities are incorrect assumptions, but for the initial calculation with the green at the same height, you do know the correct non-zero final velocity components.

 

[PeroK]

Since initial and final velocity remain the same, the ball will be in the air for the same amount of time as Part A and should land only 10 m higher than that of Part A which is 137.5 m in the y-dir and still 441.66 m in the x-dir.

What am I doing Wrong?

Imagine instead of a green, you have two areas of thin paper that the golf ball will fly straight through, leaving a hole where it would have landed. The first sheet of paper is 10m above the other, which is at ground level.

Now, think about the trajectory of the ball (do a rough diagram) and show on the diagram where the ball will go through the first sheet of paper and then where it will go through the second.

Note also that the ball will hit the ground at its final velocity. What happens to the ball after it hits the ground is not predicted by your kinematic equations for flight. It could bounce several times, it could land in a bunker and stop quickly or it could even hit a low wall and bounce backwards. Only the initial flight of the ball is governed by your kinematic equations for a projectile.