Verified answer

dv=(vi)v * tup + 1/2a(tup)^2

tup = (vf-vi/g)v

tup = .897 seconds

(vi)v = 15cos(54) = 8.8 m/s

dv = (8.8 m/s)(.897sec) + 1/2(-9.81m/s^2)(.805sec^2)

dv=3.9 ft

Max height is 3.9 meters

you can also test out this calculation by using the free fall equation which is d=1/2at^2 and it works out nicely......

d= 1/2at^2

-3.945m=1/2(-9.81m/s^2)(.805sec^2)

-3.945 m = -3.946m

slight error due to rounding

the person below me is wrong lol..... d finds the distance from the inital starting point to the highest point the ball reaches.... not the vertical or max height.

To find the X Component of the initial velocity you take the sin of the angle and multiply by initial velocity.

sin(36) = .5877852523

(.5877852523)(18) = (10.58013454)

Then using V = (t)(g) you can find the time it will take to reach max height.

(10.58013454) = (t)(9.8)

t = 1.079605565s

The using D = (1/2)(g)(t^2) you can find D which would be the height

D = (1/2)(9.8)(1.165548177)

D = 5.711186067m

So the height is about 6 meters

6 meters