Question

1. A basketball player passes the ball to his teammate at an angle of 40degrees above the horizontal. The ball is released 1.5m above the floor at a speed of 5m/s. Find:

a) the maximum height reached by the ball from the floor

b) the horizontal required to reach the maximum height

c) the horizontal distance the ball traveled during this time



2. A book slides off a horizontal table 1m high with a speed of 2m/s. Find:

a) the horizontal distance of the book from the table when it strikes the floor

b) the time of travel and the horizontal distance traveled if the initial speed is doubled.

Answer 1

1. Vo = 5 m/s @ 40 Deg.
Xo = 5 cos40 = 3.83 m/s.
Yo = 5 sin40 = 3.21 m/s.

a. Yf^2 - Yoo^2 + 2g*h,
h = (Yf^2 - Yo^2) / 2g,
h = 1.5 + (0 - (3.21)^2) / -19.6=2.03m.

b.

c.Tr = (Yf - Yo) / g,
Tr = (0 - 3.21) / -9.8 = 0.328 s. = Rise time or time to reach max. ht.

Dx = X0 * Tr = 3.83 * 0.328 = 1.26 m.

2. d = Vo*t + 0.5g*t^2 = 1 m.
0 + 4.9t^2 = 1,
t^2 = 0.204,
t = Tf = 0.452 s. = Fall time or time in flight.

a. Dx = Xo * Tf = 2 * 0.452 = 0.904 m.

b. Time of travel remains the same.
Dx = 4 * 0.452 = 1.81 m.