A golfer on a level fairway hits a ball at an angle of 42* to the horizontal that travels 100 yd 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 100 yd. At what angle was the second ball hit? (Neglect air resistance.)

I bet it was 90-42.

To find the angle at which the second ball was hit, we can use the concept of projectile motion. The horizontal and vertical motions of a projectile are independent of each other.

Let's break down the given information and steps to find the solution:

1. The first ball was hit with an angle of 42 degrees to the horizontal and traveled 100 yards.
2. We know that the initial velocity of the first ball is the same as the second ball, but the angle is different.
3. Both balls traveled the same horizontal distance, indicating that the time of flight for both balls is the same.

Now let's find the initial velocities of both balls using the given information:

1. Break down the initial velocity of the first ball:
- The horizontal component of velocity remains constant throughout the motion, so the horizontal velocity is equal to the horizontal distance traveled divided by the time of flight.
- The horizontal distance traveled by the first ball is 100 yards.
- Since the time of flight is the same for both balls, we can denote it as "t".
- Therefore, the horizontal velocity of the first ball is (100 yards / t).

2. Now, let's analyze the vertical component of velocity for the first ball:
- We know that the vertical displacement is zero since the ball hits the ground at the same level it was launched (on level fairway).
- The vertical displacement formula is given by:
d = v₀y * t + (1/2) * g * t²
where:
d = vertical displacement (zero in this case)
v₀y = initial vertical velocity of the first ball
g = acceleration due to gravity (assumed to be 32 feet/s² since we're using yards)
t = time of flight (unknown but the same for both balls)
- With the given initial angle of 42 degrees, we can break down the initial velocity into vertical and horizontal components:
v₀y = v₀ * sin(42°) [vertical component]
v₀x = v₀ * cos(42°) [horizontal component]
where:
v₀ = initial velocity (common for both balls)
sin(42°) = vertical component
cos(42°) = horizontal component

3. Using the above equation (d = v₀y * t + (1/2) * g * t²), since the vertical displacement is zero, the equation simplifies to:
0 = v₀ * sin(42°) * t + (1/2) * g * t²

Now, we can analyze the second ball. We know its initial velocity is the same as the first ball, but the angle is different. Let's denote the angle of the second ball as "θ".

4. Calculate the vertical component of velocity for the second ball using the angle θ:
v₀y2 = v₀ * sin(θ)

5. Substitute the equations for the vertical components of the first and second balls:
v₀ * sin(42°) * t = v₀ * sin(θ) * t

6. Simplify by canceling the common terms:
sin(42°) = sin(θ)

7. Rearrange the equation to solve for θ:
θ = arcsin(sin(42°))

8. Calculate the value of θ using the trigonometric identity:
θ = 42°

Therefore, the second ball was hit at an angle of 42 degrees to the horizontal, just like the first ball.