- #1
jutman1776
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Homework help needed!
the problem is stated as follows:
A playground is on the flat roof of a city school, 5.1 m above the street below (see figure). The vertical wall of the building is h = 6.20 m high, to form a 1.1-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 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. (18.11 m/s)
(b) Find the vertical distance by which the ball clears the wall. (1.89)
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
the answer to A) i got by V=d/t or v= 24 m/2.2 s and got 10.90 m/s then I divided 10.90/cos(53) and got the initial velocity of 18.11 m/s.
B) i did displacement of y=voyt-1/2gt2 or Δy= (14.46 m/s)(2.2 s)-1/2(9.80 m/s2)(2.2)2 and got 8.09 m/s then i subtracted that from the height of the building so 8.09 m/s-6.20 m= 1.89 m
C) this is where I am stumped i know you can use the quadratic equation but would like to learn the way in which you can use two different equations to bypass that. I just have no clue where to start.
thanks for the help in advance!,
Justin
the problem is stated as follows:
A playground is on the flat roof of a city school, 5.1 m above the street below (see figure). The vertical wall of the building is h = 6.20 m high, to form a 1.1-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 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. (18.11 m/s)
(b) Find the vertical distance by which the ball clears the wall. (1.89)
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
the answer to A) i got by V=d/t or v= 24 m/2.2 s and got 10.90 m/s then I divided 10.90/cos(53) and got the initial velocity of 18.11 m/s.
B) i did displacement of y=voyt-1/2gt2 or Δy= (14.46 m/s)(2.2 s)-1/2(9.80 m/s2)(2.2)2 and got 8.09 m/s then i subtracted that from the height of the building so 8.09 m/s-6.20 m= 1.89 m
C) this is where I am stumped i know you can use the quadratic equation but would like to learn the way in which you can use two different equations to bypass that. I just have no clue where to start.
thanks for the help in advance!,
Justin