| Tg General Gr: I The Openir l-, Timeout ale vh HW4:2 × LI Buy ATV Ti\' e An El
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| Tg General Gr: I The Openir l-, Timeout ale vh HW4:2 × LI Buy ATV Ti' e An Electror | Answers Tr l e An Electror | Maverick Trl + LTX webassign.n udent/Assignment-Responses/submit?dep-17215307 5. ÷ -B points SerPOP5 3.P.021 My Notes Ask Your Teacher A playground is on the flat roof of a city school, 6 m above the street below (see figure). The vertical wall of the building is h 7.20 m high, forming a 12-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 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.) m/s (b) Find the vertical distance by which the ball clears the wall (c) Find the horizontal distance from the wall to the point on the roof where the ball lands | 12:37 PM Type here to search 9/23/20173Explanation / Answer
set the equation of motion in the vertical as y=0 at the surface of the playground
for the ball
y(t) = -6 + v0*sin(53)*t - 0.5*9.8*t^2
we are given that
24 = v0*cos(53)*2.20
solve for v0 = 18.13 m/s
check with y(t) to see if the ball is above the wall
y(2.2) = -6 + 18.13*sin(53)*2.2 - 0.5*9.81*2.2^2
y(2.2) = 2.11 m, which is 0.91 m above the wall.
Let's find t for when y(t)=0
There will be two roots, use the greater root. The smaller value of t is when the ball reaches the plane of the playground on the ascent.
t = 2.46 seconds
In that time the ball has traveled
x(2.46) = 26.84 m
or
26.84 - 24 = 2.84 from the wall
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