Answer :
Answer:
The projectile will hit the ground after 38.8 seconds.
Step-by-step explanation:
You know that the height s (t) in feet of the projectile after t seconds is given by the function
s(t) = - 16t² + 600t + 800.
To know the time at which the projectile will hit the ground, you must replace s(t) by height 0:
0=- 16t² + 600t + 800
This is a quadratic function of the form ax2 + bx + c, where a= -16, b=600 and c=800.
To solve a quadratic function, you must apply the expression:
[tex]x1,x2=\frac{-b+-\sqrt{b^{2}-4*a*c } }{2*a}[/tex]
In this case:
[tex]x1,x2=\frac{-600+-\sqrt{600^{2}-4*(-16)*800 } }{2*(-16)}[/tex]
Solving:
[tex]x1,x2=\frac{-600+-\sqrt{360000+51200 } }{-32}[/tex]
[tex]x1,x2=\frac{-600+-\sqrt{411200 } }{-32}[/tex]
[tex]x1,x2=\frac{-600+-641.25 }{-32}[/tex]
So:
[tex]x1=\frac{-600+641.25 }{-32}[/tex]
[tex]x1=\frac{41.25 }{-32}[/tex]
x1= -1.289 seconds
and
[tex]x2=\frac{-600-641.25 }{-32}[/tex]
[tex]x2=\frac{-1241.25 }{-32}[/tex]
x2= 38.789 ≅ 38.8 seconds
Time cannot be negative. So, the projectile will hit the ground after 38.8 seconds.