Asked by Sunny
A projectile is launched upward from the ground at an initial velocity of 400 feet per second. During it’s upward flight, the projectile passes a hot air balloon floating at a constant height of 2400 feet. How many seconds later will the projectile pass the hot air balloon again on the way down?
Answers
Answered by
Steve
you want to solve for t in
400t - 16t^2 = 2400
Then find the difference in the two times.
400t - 16t^2 = 2400
Then find the difference in the two times.
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