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of the hnched at an angle aigh forming a 1.30-m-high railing around tfie playgro

ID: 1875002 • Letter: O

Question

of the hnched at an angle aigh forming a 1.30-m-high railing around tfie playground. A ball of the building of a school, 5.80 m above the street below. The vertical wall of = 53.0° above the horizontal at a point d = 24.0m from the base the building is h 7.10 m was laun 1.30-m-high at an angle wall. The e ball takes 2.20s to reach a point vertically above the wall. (a) (b) (6 points) Find the speed at which the ball was launched. nd the vertical distance by which the ball clears the wall. 1,00m nd the horizontal distance from the wall to the point on the roof where the (c) (7points) Fi ball lands.

Explanation / Answer

given h= 5.8 m

H = 7.1 m

theta = 53 deg

d = 24 m

t = 2.2 s

a. let launch speed be v

then

d = vcos(theta)*t

hence

v = 18.12 m/s

b. vertical distance by which ball clears the wall = w

then

y = vsin(theta)*2.2 - 0.5g2.2^2

y = 8.1 m

w = y = H = 1.033 m

c. let the horizontal distance be l

then

x = vcos(theta)*t'

5.8 = vsin(theta)*t' - 0.5*g*t'^2

4.9t'^2 - 14.47127*t' + 5.8 = 0

hence

t' = 2.472562705 s

hence

x = 26.9630199 m

hecne l = x - d= 2.9630 m