Question

A person throws a ball upward into the air with an initial velocity of 15.0 m/s.
Calculate the following:
a.) how high it goes.

b.) how long the ball is in the air before it comes back to his hand.

c.) at what time(s) will the ball pass a point 8.00 m.
If you got multiples times, are both solutions valid? Why or why not?

Answer 1

a. h = (Vf^3 - Vo^2) / 2g,
h = (0 - (15)^2) / -19.6 = 11.5m.

b. t(up) = (Vf - Vo) / g,
t(up) = (0 - 15) / -9.8 = 1.53s.

t(dn) = t(up) = 1.53s.

T = t(up) + t(dn) = 1.53 + 1.53 = 3.06s. = Time in flight.

c. d = Vot + gt^2 = 8m,
15t -4.9t^2 = 8,
-4.9t^2 + 15t -8 = 0,
Use Quadratic Formula to find t:
t = 0.69s.