Help Needed: Solving Golf Ball Kinetic Energy Problem

AI Thread Summary
The discussion revolves around solving a physics problem involving a golf ball's kinetic energy and speed at different heights. The golf ball, weighing 47.0 g and initially traveling at 52.0 m/s, reaches a height of 23.2 m. At this maximum height, its velocity is zero, resulting in zero kinetic energy. The participants emphasize the conservation of mechanical energy, stating that potential energy at the highest point can be used to determine the kinetic energy and speed at lower heights. Calculations involve using the initial speed and height to find the required values, highlighting the importance of understanding energy conservation principles in this context.
miranda82
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I have worked out this problem multiple times and i am still getting it wrong, HELP!

I need some help with this, anyone taking a shot is worth it, i would greatly apprieciate

A 47.0 g golf ball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 23.2 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
J
(b) What is its speed when it is 9.0 m below its highest point?
m/s
it a/a/a/p if anyone can figure it out. Thank you, M.
 
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You should show us what you've done so far. Were you able to determine the components of the velocity at the time when it left the tee?
 
Hello, miranda82!
Well, the answers are:
a)When the ball reachs maximum height(in this case, 23.2 m) its velocity is null(zero), because it breaks instantly to change its direction. Consequently, its kinetic energy is null(zero) too.

b)How there is no air resistance or any friction force, but only weight force(which is resultant force) which is a conservative force, so , the mechanics energy is conservatived. Therefore,
mechanical energy at the highest point:mgH(according to (a))
mechanical energy at a point 9.0 m below to highest point:mgh+(1/2)*mv^2
therefore,
mgh+(1/2)*mv^2 = mgH, is equivalent to [2g(H-h)]^1/2=v, which is the velocity that you're finding.
PS.:H:maximum height
h:height 9.0m below to highest point.
Now, it's only substitute the datas and good calculus!
 
kastarov said:
a)When the ball reachs maximum height(in this case, 23.2 m) its velocity is null(zero), because it breaks instantly to change its direction. Consequently, its kinetic energy is null(zero) too.
Only if it's going straight up.

I haven't done the calculations, but I assume that if the ball had been going straight up, it would have reached a higher altitude than that. You have to use the altitude given to figure out how much of the initial velocity was in the "up" direction and how much was in the "forward" direction.
 
You are given the initial speed and so can calculate the kinetic energy as the ball leaves the tee. You know the height the ball reaches so you can calculate the potential energy (relative to the ground) there. Since, as Fredrik said, energy is conserved, the kinetic energy at the highest point is The kinetic energy on the ground minus the potential energy at the highest point. You can find the speed of the ball at the highest point from that.

For part b, just figure out how high the ball is when it is "9.0 m below 23.2 meters" and do the same as in (a).
 
HallsofIvy said:
Since, as Fredrik said, energy is conserved...
I didn't say that, but I could have said it. :smile: I actually had another solution in mind, but yours is easier.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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