Asked by ashley

A ball is thrown vertically upward with a speed of +13.0 m/s.
(a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after it reaches its highest point?
(d) What is its velocity when it returns to the level from which it started?

Answers

Answered by Henry
a. h = (Vf^2 - Vo^2) / 2g,
h = (0 - (13)^2) / -19.6 = 8.62m.

b. t(up) = (Vf - Vo) / g,
t(up) = (0 - 13) / -9.8 = 1.33s.

c. d = Vo*t + 4.9t^2 = 8.62,
13t + 4.9t^2 = 8.62,
4.9t^2 + 13t - 8.62 = 0Use Quadratic Formula to find t:
t = 0.55s.

d. Vf^2 = Vo^2 + 2gd,
Vf^2 = 0 + 2*9.8*8.62 = 168.952,
Vf = 13m/s. = Same as initial velocity
going up.


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