A playground is on the flat roof of a city school, 5.7 m above the street below

Question

A playground is on the flat roof of a city school, 5.7 m above the street below (see figure). The vertical wall of the building is h = 7.10 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 theta = 53.0 Degree 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.) 18.13 m/s (b) Find the vertical distance by which the ball clears the wall. 1 m (c) Find the horizontal distance from the wall to the point on the roof where the ball lands. m

Explanation / Answer

a)

vx = d/t = 24/2.2 = 10.9091 m/s

v = vx/cos = 10.9091/cos53 = 18.127 = 18.13 m/s

b)

H = -0.5 g t^2 + v sin t = -0.5*9.8*2.2*2.2 + 18.127*2.2*0.798636 = 8.1331 m

h' = 8.1331 - 7.10 = 1.0331 = 1.03 m

c)

H' = 1.0331 + 1.4 = 2.4331 m

v'y = vy - g t = 18.127*0.798636 - 9.81*2.2 = -7.1051 m/s

H' = -0.5 g t^2 + v'y t

2.4331 = -0.5*9.8*t^2 - 7.1051 t

>>> t = 0.5554 s

>>> d' = 0.5554 * 10.9091 = 6.06 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.