i noticed you helped a lot of people with their questions with physics ? can you hep me out please ?


In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say 2.00m long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, ) is more difficult than from a downhill lie.

part a)To see why, assume that on a particular green the ball decelerates constantly at 1.6m/s^2 going downhill, and constantly at 2.5m/s^2 going uphill. Suppose we have an uphill lie 7.0m from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range 2.00m short to 2.00m long of the cup.

part b)Do the same for a downhill lie 7.0m from the cup.

First consider the "uphill lie", in which the ball is being putted down the hill. Choose

x (o) to be the ball's original location, and the direction of the ball's travel as the positive direction. The final velocity of the ball is v(f) = 0 m/s , the acceleration of the ball is a = - 1.6 m/s^2, and the displacement of the ball will be 7±2 m, therefore, 5 m for the first case and 9 m for the second case
From
s={v(f)^2 –v(i)^2}/2a
initial velocity of the ball will be
v(i) = sqrt(2as) = sqrt(2•1.6•5) = 4 m/s
v(i) = sqrt(2as) = sqrt(2•1.6•9) = 5.4 m/s.
The range of acceptable velocities for the uphill lie is 4 m s to 5.4 m s , with a spread of 1.4 m/s.

Now consider the "downhill lie", in which the ball is being putted up the hill. Use a very similar setup for the problem, with the basic difference being that the acceleration of the ball is now a = - 2.5 m/s^2.
initial velocity of the ball
v(i) = sqrt(2as) = sqrt(2•2.5•5) = 5 m/s
v(i) = sqrt(2as) = sqrt(2•2.5•9) = 6.7m/s.
The range of acceptable velocities for the uphill lie is 5 m s to 6.7 m s , with a spread of 1.7 m/s.

Because the range of acceptable velocities is smaller for putting down the hill, more control in
putting is necessary, and so the downhill putt is more difficult.

thank you so much !

sorry wrong post

Sure, I can help you with that!

To solve this problem, we need to use the equations of motion to calculate the allowable range of initial velocities for both uphill and downhill lies.

Part a) Uphill lie:
Given:
Acceleration (a) uphill = -2.5 m/s^2 (negative sign indicates deceleration)
Distance to the cup (d) = 7.0 m
Allowable range of stopping distances = 2.00 m (short) to 2.00 m (long)

We can use the following equation of motion to solve for the initial velocity (v) uphill:
d = (v^2 - u^2) / (2a)
where d is the distance, v is the final velocity (0 in this case), u is the initial velocity, and a is the acceleration.

Rearranging the equation, we get:
u = sqrt(v^2 - 2ad)

Substituting the given values, we have:
u = sqrt(0 - 2(-7.0)(-2.5))
u = sqrt(35)

Therefore, the allowable range of initial velocities for an uphill lie is sqrt(35) m/s.

Part b) Downhill lie:
Given:
Acceleration (a) downhill = -1.6 m/s^2 (negative sign indicates deceleration)
Distance to the cup (d) = 7.0 m
Allowable range of stopping distances = 2.00 m (short) to 2.00 m (long)

Using the same equation of motion as before:
u = sqrt(v^2 - 2ad)

Substituting the given values, we have:
u = sqrt(0 - 2(7.0)(-1.6))
u = sqrt(22.4)

Therefore, the allowable range of initial velocities for a downhill lie is sqrt(22.4) m/s.

In summary:
a) For an uphill lie, the allowable range of initial velocities is sqrt(35) m/s.
b) For a downhill lie, the allowable range of initial velocities is sqrt(22.4) m/s.