Question
A ball is thrown upward with an initial velocity of 4 ft/sec from a height of 10 feet. What is the maximum height that the ball will reach?
h(t)=-16t^2+v0t+s0
Please help im lost
h(t)=-16t^2+v0t+s0
Please help im lost
Answers
Steve
plug in your numbers:
h(t) = -16t^2 + 4t + 10
Recall from your Algebra I that this is just a parabola, with vertex at x = -b/2a, or in this case, x = 1/8
h(1/8) = 41/4
This is no time to forget your earlier math!
h(t) = -16t^2 + 4t + 10
Recall from your Algebra I that this is just a parabola, with vertex at x = -b/2a, or in this case, x = 1/8
h(1/8) = 41/4
This is no time to forget your earlier math!
Ryan
Lol thank you i don't know why i was confused when it was that simple