a ball is thrown straight upward with some unknow velocity. The ball reaches a maximum height above the launch point of 32m. how fast is the ball movin when it is 8 m above the launch point
3 answers
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(a ball is thrown straight upward with some unknow velocity. The ball reaches a maximum height above the launch point of 32m. how fast is the ball movin when it is 8 m above the launch point )
v = Vi - g t
0 = Vi - 9.8 t where t is time to stop at top
so
Vi = 9.8 t
32 = 0 + Vi t - 4.9 t^2
32 = 9.8 t^2 - 4.9 t^2
t^2 = 6.53
t = 2.56 seconds to top
so
Vi = 9.8*2.56 = 25 m/s
8 = 0 + 25 t - 4.9 t^2 where ti is time at 8 m
4.9 t^2 -25 t + 8 = 0
t^2 - 5.11 t + 1.63 = 0
t = [ 5.11 +/- sqrt(26.11-6.52) ]/2
= [ 5.11 +/- 4.43]/2
small answer is on the way up, use it
t = .34 seconds
v = 25 - 9.8*.34 = 21.7 m/s
0 = Vi - 9.8 t where t is time to stop at top
so
Vi = 9.8 t
32 = 0 + Vi t - 4.9 t^2
32 = 9.8 t^2 - 4.9 t^2
t^2 = 6.53
t = 2.56 seconds to top
so
Vi = 9.8*2.56 = 25 m/s
8 = 0 + 25 t - 4.9 t^2 where ti is time at 8 m
4.9 t^2 -25 t + 8 = 0
t^2 - 5.11 t + 1.63 = 0
t = [ 5.11 +/- sqrt(26.11-6.52) ]/2
= [ 5.11 +/- 4.43]/2
small answer is on the way up, use it
t = .34 seconds
v = 25 - 9.8*.34 = 21.7 m/s