Question

Suppose a football player kicks a ball and gives it an initial upward velocity of 47 ft/s. Use the formula h(t)= -16t^2+47t+3 where h(t) is the height and t is the time.
A. What is the starting height of the football?
B. How high did the football go?
C. How long did it take for the football to go to its highest point?
D. If no one catches the football, how long will it be in the air?

Answer 1

Answer:

Step-by-step explanation:

a. when t=0,h=3ft

$h(t)=-16 (t^2-\frac{47}{16}x+(\frac{47}{32} )^2)+\frac{47^2}{32^2} *16+3\\=-16(t-\frac{47}{32})^2+\frac{2209}{64}+3\\=-16(t-\frac{47}{32})^2+\frac{2209+192}{64}\\when~ t=\frac{47}{32}\\h=\frac{2301}{64}\\or~v^2-u^2=2gh\\\\0^2-47^2=2*(-32)h\\or h=\frac{47^2}{64}=\frac{2209}{64} \\height from the ground=\frac{2209}{64} +3=\frac{2301}{64}$

c.t=47/32 sec

d.

time taken to fall 2301/64 ft=0*t+\frac{1}{2}*32*t^2

t^2=\frac{2301}{64*16}

t=\frac{\sqrt{2301}}{8*4}=\frac {47.97}{32}

total ~time=\frac{47+47.97}{32}=\frac{94.97}{32} sec