Asked by Kay
with what minimum speed must a ball be thrown straight up in order to reach a height of 34 meters above the launch position?
Answers
Answered by
Damon
v = Vi - 9.81 t
at top v = 0
so
t = Vi/9.81 at top
34 = 0 + Vi t - (9.81/2) t^2
34 = Vi^2/9.81 - Vi^2 /(2*9.81)
34 = Vi^2/(2*9.81)
Vi = 25.8 m/s
at top v = 0
so
t = Vi/9.81 at top
34 = 0 + Vi t - (9.81/2) t^2
34 = Vi^2/9.81 - Vi^2 /(2*9.81)
34 = Vi^2/(2*9.81)
Vi = 25.8 m/s
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