Projectile question. Please help....

A golf ball is struck 50 degree with respect to the horizontal at a velocity of 23 ms^-1. The golf course is perfectly level and the ball stops instantaneously as it hits the ground.

a) Determine the horizontal distance traveled by the golf ball

b) Determine the maximum height attained by the golf ball

c) If the golf ball is to travel a distance of 30m, using the same initial trajectory, determine the required initial velocity

Please leave your working as well. I was trying to do this question for hours.....

Veocity, v = 23 m/s,
Angle, Ɵ = 50 degrees

Horizontal-velocity, vh = v.cosƟ = 23.cos50 = 14.8 m/s,

Vertical velocity, vv = 23.sin50 = 17.6 m/s,

Time of flight, t = 2*vv/g = 2*17.6/9.8 = 3.6 seconds,

a)
Horizontal distance = vh*t = 14.8*3.6 = 53.22 meters >==================< ANSWER

b)
Max. height, h = (vv)^2/2g = (17.6)^2/2*9.8 = 15.8 meters >=============< ANSWER

c)
Let,
initial velocity, = v m/sec,
SO,
time of flight, t1 = 2vsin(50)/9.8 seconds,

distance = vcos(50)*t1 = vcos(50)*2vsin(50)/9.8 = 30 meters,
OR,
v^2 = 30*9.8/2sin(50).cos(50) = 294/sin(100) = 298.54,
OR,
v = sqrt(298.54) = 17.28 m/sec >==============================< ANSWER