Please help, I have tried working this out numerous times and keep getting stuck
ID: 2002753 • Letter: P
Question
Please help, I have tried working this out numerous times and keep getting stuck! Thank you.
A playground is on the flat roof of a city school, hb = 6.50 m above the street below (see figure). The vertical wall of the building is h = 7.70 m high, to form a 1.2-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 = 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.
Explanation / Answer
Ans: The question is not clear and diagram is not present.
But I think you knoe the distance between the passerby and the top point of the wall and it is =(7.7^2 + 24 ^2)(1/2) m=25.2m. So the ball have to travel 25.2m in 2.20 sec.
if the passer by through it with a initial velocity u. then
25.2=u*t +0.5*g*t^2 ; solving : u= 0.674 m/sec.
For proper explanation and answer you need to be more clear and please attach the figure.