posted by anonymous .
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 1.5-m-high 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.
(b) Find the vertical distance by which the ball clears the wall.
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
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
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)
y = 8.14
c) Use the same equation to solve for t when y = 5.8 m
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:
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.