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 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 p

Answer 1

The diagram somehow got shifted towards right..the part that got missing is only labeling of heights.

Now, can you post your attempt for the question ??

Answer 2

where is the answer?

Answer 3

Why don't you show me your attempt first ??
Whatever equations you made..or anything..

Answer 4

here is the diagram

Answer 5

Look, if you are just looking for an answer to your question; then I can't help you...its also against the Code of Conduct of this website.
http://openstudy.com/code-of-conduct

Answer 6

ok

Answer 7

i get first the initial velocity

Answer 8

Vi=24/2.2cos53

Answer 9

Vi=18.1m/s

Answer 10

" (c) Find the horizontal distance from the wall to the p"
Can you complete the question ??

Answer 11

after that i don't know what to do

Answer 12

. (c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

Answer 13

Okay..

The calculation you did for finding velocity is wrong..d < range of the projectile.
Hence you CANNOT write : Vcos(53)*2.2 = 24

Answer 14

what is the correct?

Answer 15

just give me the formulas

Answer 16

Can you write the equations for the motion in x and y direction ??

Answer 17

i don't know that

Answer 18

just give to me the formulas

Answer 19

The usual equations are -
y = Vsin(53)t - (1/2)gt^2
x = Vcos(53)t

Vy = Vsin(53) - gt
Vx = Vcos(53)

Find the time when y = 7 ; you'll get two values of time ; let the bigger one be T
Then, Vcos(53)T = 24 ; you can find V from here

Next; find time when x = 24; substitute it in y = Vsin(53)t - (1/2)gt^2 to get ; then (y - 7) is the answer for part (b)

Answer for part (c) is ( Vcos(53)*2.2 - 24)

Answer 20

what is V stands for?

Answer 21

V is the magnitude of initial velocity vector.