A body is thrown straight up with initial velocity 5 feet per second from a height of 40 feet. After how many seconds will it hit the ground? What will be its maximum height?
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keemariee
ANSWERS: 1
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Formula: height of ball = 40 + 5 t - (1/2) g t^2 Set this to zero and solve for t to find when it hits the ground. One way to find the maximum height is to complete the square. Suppose g=32 Then h = 40 + 5 t - 16 t^2 h = - 16 (t^2 - (5/16) t - (5/2)) complete the square: h = - 16 ( (t-(5/32))^2 - (25/1024) - (5/2)) h = - 16 ( (t-(5/32))^2 - (2585/1024)) h = 2585/64 - 16 ((t-(5/32))^2 clearly the maximum value occurs when the square is zero.
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