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Time to reach the top of the arc = (25/9.8) seconds.

Time to return to the height from which it was launched = another (25/9.8) seconds.

Time to fall the additional 1.5 meters:

D = V0 t + 1/2 g t2

1.5 = (25) t + 4.9 t2

4.9 t2 + 25 t - 1.5 = 0

t = (1/9.8) [ -25 +/- sqrt(625 + 6 x 4.9) ] = (1/9.8) [ -25 +/- sqrt(654.4) ] = 0.0593 sec

Total time = (50/9.8) + (0.0593) = 5.695 seconds.

Assumes:

-- (ground elevation at launch point) - (ground elevation at landing point) = 0

-- no air resistance

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Q: A javelin thrower releases a javelin 1.5 meters from the ground with an intital vertical velocity of 25 meters per second How long will it take the javelin to hit the ground?
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