13:56 LTE a webassign.net check your calculations. m A playground is on the flat

Question

13:56 LTE a webassign.net check your calculations. m A playground is on the flat roof of a city school, -6.30 m above the street below (see figure). The vertical wall of the building is h-7.50 m high, to form a 1.2-m-high railling around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of 8-53.0° above the horizontal at a point d-24.0m 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 (b) Find the vertical distance by which the ball clears the wall. Your response differs from the correct answer by more than 10% Double check your calculations. m (c) Find the horizontal distance from the wall to the point on the roof where the ball lands. Your response differs from the correct answer by more than 10%. Double check your calculations. m Need Help? Home My AssignmentsExension Request

Explanation / Answer

a)

h= 7.50m, h0 =6.30m, d=24.0m, t=2.20s, ?=53,0 deg

d= vix*t

d= vi*cos?*t

24.0=vi*cos53*2.2vi=18.127 m/s

b)

y=viy*t-1/2*gt^2

y=vi*sin?*t-1/2*gt^2

y=18.127*2.2-1/2*9.8*2.2^2

y=16.16m

This is a total vertical distance reached by the ball in 2.2s.

Vertical distance above the wall = 16.16 – 7.50 = 8.66m

c)

Let us find vertical velocity at t=2.2s

vfy=viy-gt

vfy=18.127*sin53-9.8*2.2

vfy=-7.08m/s

vfx=vix=18.127*cos53=10.91 m/s

Let us find time to return the ball on the roof. So ball have to lower the vertical distance to the roof

=16.16-6.30 = 9.86m

Use projectile motion eqn,

?y=vfy*t-1/2gt^2

-9.86=-7.08*t-1/2*9.8*t^2

t=0.87s

Horizontal distance from the wall = x

x=vix*t

x= 10.91*0.87

x= 9.49 m

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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