Posted by **Tomtom** on Sunday, October 16, 2011 at 5:17pm.

You stick your arm over the edge of the Grand Canyon, and throw a ball vertically upwards with a velocity of 15 m/s. Compared to the heighht that you throw the ball, what height is the ball after 6 seconds?

- Physics -
**Henry**, Monday, October 17, 2011 at 8:27pm
t(up) = (Vf - Vo) / g,

t(up) = (0 - 15) / -9.8 = 1.53s. to reach max. ht.

hmax = (Vf^2 - Vo^2) / 2g,

hmax = (0 - (15)^2) / -19.6 = 11.5m.

above launching point.

h = Vo*t + 0.5g*t^2,

h = 15*6 - 4.9*6^2 = -86.4m = The distance below max. ht.

Free-fall distance=-86.4 - 11.5 = -98m

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