A golfer on a level fairway hits a ball at an angle of 37 degrees to the horizontal,and it travels 100yd before striking the ground.He then hits another ball from the same spot with the same speed,but at a different angle. This ball also travels 100yd. At what angle was the second ball hit?

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A golfer on a level fairway hits a ball at an angle of 37 degrees to the horizontal,and it travels 100yd before striking the ground.He then hits another ball from the same spot with the same speed,but at a different angle. This ball also travels 100yd. At what angle was the second ball hit?


Interesting question.

This seems to defy the laws of physics unless I am missing something here.

Anyone else got a solution?


Tut

It's not impossible Tut, using the correct equations, it is simple actually :)

There's a speed with which the ball is hit: Let's call it v.
There's an angle, let's call it theta, \theta

Then we start on the equations.

The net vertical displacement (vertical distance from the finish point to the starting point) is zero... but is given by a kinematic equation using the vertical component of the speed of the ball.

s = ut + \frac12 at^2

0 = v\sin\theta t - \frac12 gt^2

And the horizontal distance is 100.

s = ut

100 = v\cos\theta t

Plug and chug 37 degrees to get two equations with two variables. (Simultaneous equations.) What is needed here is v (t is not as important, for v is needed for the next part).

Then, we come to the next part, getting this time, another angle mu, \mu

s = ut + \frac12 at^2

0 = v\sin\mu t - \frac12 gt^2

And the horizontal distance is 100.

s = ut

100 = v\cos\mu t

Plug in v, to get two equations with two variables, this time the angle and the time. Solve for the angle. :) This should give you 2 angles one of which is the angle from the first part, and the other angle is the one you're looking for. (and of course, the angle should be 0\ <\ \mu\ <\ 180^o)