acceleration is a constant 9.8 m/s^2 downward, so
v = 8 - 9.8t
max height reached when v=0 (it stops going up)
s = 4.9t^2
plug in your value of t
It takes twice that long to return to the starting point.
a 1kg ball is thrown vertically. If the ball has a velocity of 8 m/s at release, please find: the maximal height that the ball can reach and the time it took for the ball to reach the maximal height and how long does it take for the ball to return to it’s starting position
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