Question

A ball is tossed from an upper-storey window of a building.The ball is given an initial velocity of 8 m/s at an angle of 20 degrees below the horizontal.it strikes the ground 3s later .(a) How far horizontally from the base of the building does the ball strike the ground? (b)Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0m below the level of launching?

Answer 1

Vo = 8 m/s. @ 20o = Initial velocity.
Xo = 8*cos20 = 7.52 m/s = Hor. comp.
Yo = 8*sin20 = 2.74 m/s = Ver. comp.

a. d = Xo * Tf = 7.52m/s * 3s.=22.56 m.

b. h = Yo*Tf + 0.5g*Tf^2.
h = 2.74*3 + 4.9*3^2 = 52.32 m.

c. h = Yo*t + 0.5g*t^2 = 10 m.
2.74t + 4.9t^2 = 10
4.9t^2 + 2.74t - 10 = 0
Use Quad. Formula:
t = 1.18 s.

Answer 2

how did you get that 0.5g??

Answer 3

in reply to "how did you get that 0.5g??":
0.5g= (1/2)(g)=(1/2)(9.8)