math
posted by anonymous on .
A playground is on the flat roof of a city school, 5.8 m above the street below (see figure). The vertical wall of the building is h = 7.30 m high, to form a 1.5mhigh railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.
(a) Find the speed at which the ball was launched.
m/s
(b) Find the vertical distance by which the ball clears the wall.
m
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
m

a) The horizontal component of the launch speed (Vo) is Vo cos = 0.6018 V. It travels 24 m horizontally in 2.20 s. Therefore
X= X_0+V_x0(t)
24= 0+V_x(2.2)
10.91= V_0cos53
V_0= 18.13
b)to solve for the height after t = 2.2s
y = (Vo sin 53)* t  (1/2) g t^2
= ((18.13sin53)*2.2) 4.9(2.2^2)
= 31.8623.72
y = 8.14
8.147.3= .84
c) Use the same equation to solve for t when y = 5.8 m
5.8= ((18.13sin53)t)4.9t^2
= 4.9t^2+14.18t+5.8
I suggest that you graph this on your calculator to solve for t
whatever your answer is for t plug it in to this equation:
x= V_x(t)
then subtract 24.0 m from your answer which will finally give u the answer for c 
C) x= V_x(t) should actually be x= V_x * (t) * (cosine(53))
that should give the the correct answer for x
Then subtract that answer from 24. That is your final answer. 
Actually, I meant subtract 24 from x. That should give you the right answer.