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A projectile is launched into the air following the equation h(t)=-40t^2+200t, where h(t) is the height of the projectile above the ground in feet, and it is the time, in seconds. How much times elapses, in seconds, from the time the projectile is launched to the time the projectile hits the ground?

a)1

b)4

c)5

d)16

Respuesta :

Luv2Teach

Answer:

Step-by-step explanation:

If the position function is

[tex]h(t)=-40t^2+200t[/tex] and we are looking for time when the height is 0, we sub in a 0 for h(t) and solve for t:

[tex]0=-40t^2+200t[/tex] and the easiest way to do this is to factor by taking the GCF of -40t:

0 = -40t(t - 5) and by the Zero Product Property,

-40t = 0 or t - 5 = 0. Solving for t, we get

t = 0 (which is before the object is launched) and

t = 5 (which is how long it takes the object to go from the ground, up to its max height, and then back to the ground again).

Your choice is c) 5

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