A golf ball leaves the tee with a velocity of 59 fts at 60 d

A golf ball leaves the tee with a velocity of 59 ft/s at 60 degrees. If it lands on a green located 7 feet higher than the tee, what was the time of flight, and what was the horiztonal distance to the green?

Time = XXXX seconds

Distance = XXX feet

Solution

a = -g = 32..17 ft/sec^2 ; v = 59 ft/sec

horizontal component : vx = 59*cos60 = 29.5 ft/sec

vertical component : vy = 59sin60 = 51.1 ft/sec

Need to determine the time it took to land:

y = vyt + 1/2gt^2

7= 29.5*t - (32.17/2)t^2

16.08t^2 - 29.5t -7 =0

solve for t : t = 2.04 , -0.21

neglect -ve root.

So, t = 2.04 sec

Distance: : x = vx*t =  51.1*2.04 = 104.24 ft