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]

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