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Answer to Question #4361 in Mechanics | Relativity for Nate

Question #4361
A playground is on the flat roof of a city school, 6.1 m above the street below (see figure). The vertical wall of the building is h = 7.50 m high, forming a 1.4-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 θ = 53.0°; above the horizontal at a point d = 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. (Give your answer to two decimal places to reduce rounding errors in later parts.)
Expert's answer
Equation of motion of the ballx(t) =v0 cos(teta t) y(t)=v0 sin(teta t) -gt^2/2
[img width=144,height=59]file:///C:/Users/Sekventa/AppData/Local/Temp/OICE_18DAB834-6E09-410E-9CBC-2259ACB05FA0.0/msohtmlclip1/01/clip_image002.png[/img]
We know that at the moment t=2.2 the ball is 7.5 highover the ground.

7.5=2.2v0 sin(teta)-4.84g/2= 2.2v0 sin(53)-4.84g/2=1.757v0-23.716 => v0=17.77 m/s[img width=624,height=56]file:///C:/Users/Sekventa/AppData/Local/Temp/OICE_18DAB834-6E09-410E-9CBC-2259ACB05FA0.0/msohtmlclip1/01/clip_image004.png[/img]

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