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

@bakonloverk

Answer 2

help me please

Answer 3

@bakonloverk and thanks for the advice

Answer 4

@Rohitkhanna help me please

Answer 5

@Rohitkhanna i need a complete answer in question A,B and C

Answer 6

the ball has velocity in two directions-------->horizontal and vertical.
let the initial velocity of the ball be v
then,horizontal velocity(u)=v cos(53)
the man is 24 m away from the base of the wall so the ball has covered 24 m in 2.20s.the acceleration in horizontal direction is zero.
so now applying the equation------> s=ut+at^(2)/2
s=distance
a=acceleration
t=time
u=initial velocity

Answer 7

@Rohitkhanna can you please show me the whole solution? becasue i can't understand it

Answer 8

just put the values in the given equation
and u=v cos(53)
cos(53)=0.6
so u=0.6 x v
a=0
now try to solve it