Question

This question has a few parts to it. I tried but not sure if they are correct.
A boy standing on top of a building in Albany throws a water balloon up vertically. The height (h) in feet, of the water is given by the equation
h(t)=-16t^2+64t +192. where the time( in seconds) after he threw the water balloon. Answer the following. Show work and explain you answer.
a) How high is the building? h(t)=16(t^2 +4t+12)
h(t)=16(t+2)(t+6) this is as far as I can go, help please
b)What is the maximum height of the balloon?
h(t)16(t+2)(t+6)
0 t=-2 t=-6 so 0 is what I got
c)What is the value of t when the balloon hits the ground?
0=16(t+2)(t+6)
0=16(t+8t+12)
0=144t+192 t=-192/144
Please help since I tried but I don't think I did them correctly and I want to understand where I went wrong. thank you

Answer 1

a) the balloon is at the top of the building at time zero
... h = 192

b) the time for max height is at the vertex of the parabola
... on the axis of symmetry ... x = -b / 2a ... t = -64 / (2 * -16)
... plug the value of t into the equation to find max height

c) set h equal to zero, and use the quadratic formula to find t
... you want the positive solution

Answer 2

ok so the work for a)
h(0)=-16(0)^2 +64(0)+192
h(0)=192

b) h(2)=-16(2)^2+64(2)=192
h(2)=-64+128+192
h(2)=64+192 so h(2)=256

c) 0=-64plus/minus the square root64^2-4(-16)(192)/-32
o=-64plus/minus the square root of 4096+12288/-32
0=-64 plus/minus 128/-32 so it is -2 or 6 so the answer is 6
Thank you for your explanation. Hopefully these are correct now