Asked by Leanna
When a ball is thrown straight down from the top of a tall building, with initial velocity 30 ft/sec, the distance from the release point at time t in seconds is given by s(t)=16t^2+30t.
If the release point is 300 ft above ground, what is the velocity of the ball at the time it hits the ground?
I think you would have to average the feet per seconds but I'm not sure.
If the release point is 300 ft above ground, what is the velocity of the ball at the time it hits the ground?
I think you would have to average the feet per seconds but I'm not sure.
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