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 ?????

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=V_{0} tsinα - g t^{2}/2 g t^{2}/2-V_{0} tsinα+a=0

t^{2}-(2V_{0} sinα)/g t + 2a/g=0

D=2V_{0}^{2} sin^{2} α /g - 2a/g

t_{1⁄2}=( (2V_{0} sinα)/g ± √(2V_{0}^{2} sin^{2}α /g - 8a/g))/2.

We take t_{2}.

x'=V_{0} cosα ( (2V_{0} sinα)/g + √(2V_{0}^{2} sin^{2}α /g - 8a/g) ) /2 =

= 18.1cosα { ((2*18.1*sinα)/9.8 + √((18.1)^{2} sin^{2}α 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

Let a=h-1.

a=V

t

D=2V

t

We take t

x'=V

= 18.1cosα { ((2*18.1*sinα)/9.8 + √((18.1)

It is the distance from the point of throwing. Then

x = x'-d = 2.9 m

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## Comments

Mikey29.10.19, 10:45Correct, but working is very indecent and difficult to follow

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