Posted by Amy on Friday, September 6, 2013 at 2:50pm.
A ball thrown straight up from the ground passes a window 5.6 m up. An observer looking out the window sees the ball pass the window again, going down, 3.4s later.
Find the velocity with which the ball was initially thrown.

physics  Graham, Friday, September 6, 2013 at 7:20pm
Use:
s(t) = s(0) + v(0) t  ½ g t²
v(t) = v(0)  g t
9 = 9.8[m/s²]
Let the ball pass the window up at t=0[s] and down at t=3.4[s], so:
v(3.4) = v(0)
v(3.4) = v(0)  9.8×3.4[m/s]
.: v(0) = 9.8×3.4/2 [m/s]
v(0) = 16.66[m/s]
This is the velocity at the window sill.
Let the ball have been thrown at t = ŧ.
s(ŧ) = 5.3[m/s]
s(ŧ) = 16.66 ŧ  ½×9.8 ŧ²
.: 4.9 ŧ² 16.66 ŧ  5.3 = 0
ŧ = (16.66±√(16.66²+4×4.9×5.3))/(2×4.9)
ŧ = 1.7±1.993...[s]
The negative root gives the time of throw, while the positive gives the time of return.
.: ŧ ≈ 0.293...[s]
Initial velocity at time of throw:
v(ŧ) = v(0)  g ŧ
.: v(ŧ) = 1.66 + 9.8 × 0.293...
v(ŧ) = 6.57...[m/s]