Answer to Question #2773 in Mechanics | Relativity for tom

Question #2773
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 h = 7.00 m high , forming a 1-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 second to reach a point vertically above the wall .
1- Find the speed at which the ball was launched ?
2- Find the vertical distance by which the ball clears the wall ?
3- Find the horizontal distance from the wall to the point on the roof where the ball lands ?
i answer number speed = 18.1 m/s and number 2 vertical distance = 1.13 m
i need help for number 3 ?????
Expert's answer
y(t)=V_0 tsinα-g t^2/2.

Let a=h-1.

a=V0 tsinα - g t2/2 g t2/2-V0 tsinα+a=0
t2-(2V0 sinα)/g t + 2a/g=0
D=2V02 sin2 α /g - 2a/g
t1⁄2=( (2V0 sinα)/g ± √(2V02 sin2α /g - 8a/g))/2.
We take t2.
x'=V0 cosα ( (2V0 sinα)/g + √(2V02 sin2α /g - 8a/g) ) /2 =
= 18.1cosα { ((2*18.1*sinα)/9.8 + √((18.1)2 sin2α 2/9.8 - (8*6)/9.8) )/2 } = 26.9 m
It is the distance from the point of throwing. Then
x = x'-d = 2.9 m

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29.10.19, 10:45

Correct, but working is very indecent and difficult to follow

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