Question

A firecracker shoots up from a hill 160 feet high with an initial speed of 105 feet per second. Using the formula H(t) = −16t2 + vt + s, determine how long it will take the firecracker to hit the ground.

7.5 seconds

7.8 seconds

8.1 seconds

8.3 seconds

Answer 1

We have
\[H(t)=-16t^2+vt+s\]
here v= 105 feet/s
and s=160
so
\[H(t)=-16t^2+105t+160\]
here H(t) is height as function of time
we need to find the time, when the cracker hits the ground!!
Tell me what is its height when it hits the ground? @Ashley99

Answer 2

o

Answer 3

good, so h(t)=0
\[0=-16t^2+105t+160\]
Could you solve for t from this?
@Ashley99

Answer 4

i'm not sure how to use your formula
i would use
s = ut + 0.5a t^2 where s = displacement, a = acce.eration due to gravity 32ft/sec and t - time, u = iniyial velocity

-160 = 105t - 16t^2
16t^2 - 105t - 160 = 0
solve for t

Answer 5

yep using the quadratic formula (i already tried that so if you could just solve it and show the steps i would appreciate it)

Answer 6

cool, wait a minute

Answer 7

\[0=-16t^2+105t+160\]
A standard quadratic is of the form
\[ax^2+bx+c=0\]
its solution is given as
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
here b=105
a=-16
c=160
Let's plugin the values
\[x=\frac{-(105)\pm \sqrt{(105)^2-4\times (-16)\times (160)}}{2\times (-16)}\]
we get
\[x=\frac{-(105)\pm \sqrt{21265}}{2\times (-16)}\]
\[x=\frac{-(105)\pm 145.825}{2\times (-16)}\]
if we take + sign then t= -ve ( time can't be negative)
with - sign
\[x=\frac{-250.825}{-32}\]
I think you can solve now!!

Answer 8

THANKS YOU

Answer 9

You're welcome:D