A playground is on the flat roof of a city school, 4.9 m above the street below

Question

A playground is on the flat roof of a city school, 4.9 m above the street below (see figure). The vertical wall of the building is h = 6.10 m high, forming a 1.2-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

A playground is on the flat roof of a city school, 4.9 m above the street below (see figure). The vertical wall of the building is h = 6.10 m high, forming a 1.2-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 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.

Explanation / Answer

c)

above the wall

Vx = V*cos(53 degree) = 18.13*cos(53 degree) = 10.91 m/s ...(always constant)

Vy = sqrt(uy^2+2as) = sqrt((18.13*sin(53 degree))^2-2*9.81*(6.10+2.03)) = 7.08 m/s

so,

Sy = Vy*t+1/2at^2

(1.2+2.03)=7.08*t+1/2*9.81*t^2

solving

t = 0.36428 sec

so,

Sx =Vx*t = 10.91*0.36428 = 3.97 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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