Question

A juggler throws a ball straight up into the air with a speed of 11 .
With what speed would she need to throw a second ball half a second later, starting from the same position as the first, in order to hit the first ball at the top of its trajectory?

Answer 1

h = (Vf^2 - Vo^2 / 2g,
h = (0 - (11)^2) / -19.6 = 6.17m,

t(up) = (Vf - Vo) / g,
t(up) = (0 - 11) / -9.8 = 1.12s.

t = 1.12 - 0.5 = 0.62s.

Vo*t + 0.5gt^2 = 6.17m,
0.62Vo - 4.9*(0.62)^2 = 6.17,
0.62Vo - 1.88 = 6.17,
0.62Vo = 6.17 + 1.88 = 8.05,
Vo =12.98m/s.