Question

please help with:
If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 112 ft/sec, then what is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground?

Answer 1

i think you are supposed to use
\[h(t)=-16t^2+v_0t+h_0\] and in this case
\[h_0=48, v_0=112\] so you get
\[h(t)=-16t^2+112t+48\]

Answer 2

max height is at the vertex. use
\[t=-\frac{b}{2a}=-\frac{112}{2\times 16}=3.5\] to find the max height

Answer 3

velocity is given by
\[h'(t)=-32t+112\] find out when it hits the ground by solving
\[-16t^2+112t+48=0\] and then find
\[h'(t)\] for your answer

Answer 4

yea i think i had some trouble with the initial formula there, thank you!

Answer 5

yw