Asked by liberatha
the height of a ball t seconds after it is thrown into the air from the top of the building can be modelled by h(t)= -16t2+48t+64 where h(t) is height in metres.how high is the building? how high does the ball rise before starting to drop downword, and after how many seconds does the ball hit the ground?
Answers
Answered by
oobleck
first of all, your equation gives the height in feet (on earth, anyway) because in metric units g = 9.81 m/s^2
But anyway, you know that the vertex of that parabola lies at t = 48/32 = 1.5
so find h(1.5)
it hits the ground when the height is zero, so solve
-16t^2+48t+64
or, after dividing by -16,
t^2 - 3t - 4 = 0
But anyway, you know that the vertex of that parabola lies at t = 48/32 = 1.5
so find h(1.5)
it hits the ground when the height is zero, so solve
-16t^2+48t+64
or, after dividing by -16,
t^2 - 3t - 4 = 0
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.