Question

A playground is on the flat roof of a city school, 6.00 m above the street below. The vertical wall of the building is 7.00 m high, forming a 1.00 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 26.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. Find the speed at which the ball was launched. Find the vertical distance by which the ball clears the wall. Find the distance from the wall to the point on the roof

Answer 1

i need the answer now please

Answer 2

velocity = m/s
horizontal velocity = (26m)/(2.2s) =

Answer 3

then cos(53) = (horizontal velocity) / (initial velocity)
so initial velocity = ( 26/2.2 ) / (cos(53))

Answer 4

then tan(53) = (vertical velocity) / (horizontal velocity)
so vertical velocity = (horizontal velocity) tan(53)

Answer 5

wait

Answer 6

the given i give is wrong

Answer 7

I have a minute before I've got to go to school :/
and you might get a better response if you post it in the physics section
http://openstudy.com/study#/groups/Physics

Answer 8

A playground is on the flat roof of a city school, 6.00 m above the street below. The vertical wall of the building is 7.00 m high, forming a 1.00 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 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.

Answer 9

close enough
just replace everything I was saying before with a 24 instead of a 26

Answer 10

I've gtg

height = (vertical velocity)(time) + (0.5)(-9.8)(2.2 ^2)

good luck!