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.
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.
1 answer
1 answer