Asked by Tyler
A hot air balloon is rising vertically with a velocity of 4.0 feet per second. A very small ball is released from the hot air balloon at the instant when it is 640 feet above the ground. Use a(t)=−32 ft/sec2 as the acceleration due to gravity.
1. How many seconds after its release will the ball strike the ground? Round the answer to the nearest two decimal places.
2. At what velocity will it hit the ground?
1. How many seconds after its release will the ball strike the ground? Round the answer to the nearest two decimal places.
2. At what velocity will it hit the ground?
Answers
Answered by
oobleck
the height of the ball is
h(t) = 640 + 4t - 16t^2
v(t) = 4 - 9.8t
h(t) = 640 + 4t - 16t^2
v(t) = 4 - 9.8t
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