Asked by Tom
A ball is thrown in an upward direction off a platform that is 45 feet high with an initial velocity of 60 feet per second. The height, in feet, of the ball at the time t is given by
h(t)= -16t^2 + 60t + 45. The time, t, is given in seconds. Round all answers to the hundredths.
1. At what time, after the start, is the ball 45 feet high again?
2. Find the time when the ball hits the ground.
h(t)= -16t^2 + 60t + 45. The time, t, is given in seconds. Round all answers to the hundredths.
1. At what time, after the start, is the ball 45 feet high again?
2. Find the time when the ball hits the ground.
Answers
Answered by
Damon
1.
45 = 45 +60 t -16 t^2
4 t( 15-4t) = 0
t = 0 of course and t =15/4
2.
0 = 45 + 60 t -16 t^2
solve quadratic a = -16, b = 60 , c = 45
https://www.mathsisfun.com/quadratic-equation-solver.html
45 = 45 +60 t -16 t^2
4 t( 15-4t) = 0
t = 0 of course and t =15/4
2.
0 = 45 + 60 t -16 t^2
solve quadratic a = -16, b = 60 , c = 45
https://www.mathsisfun.com/quadratic-equation-solver.html
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