Question
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?
Answers
Henry
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
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