A rock is thrown upward from level ground in such a way
that the maximum height of its flight is equal to its hori-
zontal range R. (a) At what angle u is the rock thrown? (b)
In terms of its original range R, what is the range Rmax the
rock can attain if it is launched at the same speed but at the
optimal angle for maximum range? (c) What If? Would your
answer to part (a) be different if the rock is thrown with the
same speed on a different planet? Explain.

The horizontal range can be found using the range formula of a projectile on level ground. The vertical range can be found using constant acceleration kinematics. By setting the two ranges equal, we can solve for the angle.

(a) The horizontal range is:

R = v² sin(2θ) / g

Using kinematics, the vertical range is:

v² = u² + 2as

0 = (v sin θ)² + 2(-g)s

s = v² sin²θ / 2g

Setting equal, the angle is:

v² sin(2θ) / g = v² sin²θ / 2g

2 sin(2θ) = sin²θ

4 sin θ cos θ = sin²θ

4 cos θ = sin θ

4 = tan θ

θ ≈ 76.0°

(b) When thrown at angle θ from part (a), the horizontal range is:

R = v² sin(2θ) / g

When thrown at 45° for maximum range, the horizontal range is:

Rmax = v² / g

Substituting into the first equation:

R = Rmax sin(2θ)

Rmax = R / sin(2θ)

Rmax = 2.125 R

(c) The answer from part (a) is independent of the acceleration due to gravity. Therefore, the answer would not be different.