Respuesta :
Given that Martin throws a ball straight up in the air.
The equation [tex]h(t)=-16t^2+40t+5[/tex] gives the height of the ball, in feet, t seconds after martin releases it.
We need to determine the time that it takes the ball to hit the ground.
Time taken:
To determine the time 't', let us equate h(t) = 0 in the equation [tex]h(t)=-16t^2+40t+5[/tex]
Thus, we have;
[tex]0=-16t^2+40t+5[/tex]
Switch sides, we get;
[tex]-16 t^{2}+40 t+5=0[/tex]
Now, we shall solve the equation using the quadratic formula.
Thus, we have;
[tex]t=\frac{-40 \pm \sqrt{40^{2}-4(-16) 5}}{2(-16)}[/tex]
Solving, we get,
[tex]t=\frac{-40 \pm \sqrt{1600+320}}{-32}[/tex]
[tex]t=\frac{-40 \pm \sqrt{1920}}{-32}[/tex]
[tex]t=\frac{-40 \pm 8\sqrt{30}}{-32}[/tex]
[tex]t=\frac{8(-5 \pm \sqrt{30})}{-32}[/tex]
Cancelling the common terms, we get,
[tex]t=\frac{-5 \pm \sqrt{30}}{-4}[/tex]
Thus, the roots of the equation [tex]t=\frac{-5 + \sqrt{30}}{-4}[/tex] and [tex]t=\frac{-5 - \sqrt{30}}{-4}[/tex]
Simplifying the roots, we get,
[tex]t=\frac{-5 +5.477}{-4}[/tex] and [tex]t=\frac{-5 -5.477}{-4}[/tex]
[tex]t=-0.119[/tex] and [tex]t=2.619[/tex]
Since, t cannot take negative values, then [tex]t=2.619[/tex]
Hence, It takes 2.6 seconds for the ball to hit the ground.
