physics
posted by Amy .
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.

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]
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