Question

A golfer gives a ball a maximum initial speed of 37.0 m/s.
1-What is the highest tree the ball could clear on its way to the longest possible hole-in-one?

Answer 1

I do not know if you have derived the max range angle of 45 degrees. Therefore we will have to do that first.
A = launch angle up from horizontal
S = initial speed (37)
Vi = initial speed up = S sin A
U = horizontal speed = S cos A
t = time at max height = .5 total time in air so time in air = 2 t
r = range

V = Vi - g t = S sin A - g t
at max height V = 0 so g t = S sin A
and t = (S/g) sin A

h = Vi t -.5 g t^2 = S sin A t - .5 g t^2
h = S sin A(S/g)sin A = (S^2/g)sin^2 A

r = Ui 2 t = 2 S t cos A
r = 2 S (S/g) sin A cos A
so the question is: when is sin A cos A maximum?
at maximum dr/dA = 0
dr/dA = (2 S^2/g)[ -sin^2 A + cos^2 A]

Sin A = cos A when A = 45 degrees
then finally use the height equation
h = S^2/g sin^A
sin A = sqrt 2/2 so sin^2 A = 1/2
h = (37^2/9.8).5 = 69.8 m