Question

An object is launched from a platform. Its height (in meters), x seconds after the launch, is modeled by: h(x)=-5(x-4)^2+180h What is the height of the object at the time of launch?

Answer 1

Answer:

h(x) = 0

Step-by-step explanation:

An object is launched from a platform. Its height, x seconds after the launch, is modeled by :

$h(x)=-5(x-4)^2+180$ ....(1)

It is required to find the the height of the object at the time of launch. At that point, h(x) = 0

$-5(x-4)^2+180=0$

It can be written as :

$-5(x-4)^2=-180\\\\5(x-4)^2=180$

Dividing both sides by 5 i.e.

$(x-4)^2=36$

We know that, $\sqrt{36}=6$

So,

$x-4=\pm 6$

Case 1,

$x-4=+6\\\\x=10\ s$

Case 2,

$x-4=-6\\\\x=-2\ s$, time cannot be negative

Put t = 10 s in equation (1)

So,

$h(x)=-5(10-4)^2+180\\\\h(x)=0$

So, the height of the object is equal to 0 as it was at ground.

Learn more,

Projectile motion

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Answer 2

Answer:45

Step-by-step explanation:

H(0)=-5(0+1)(0-9)

= -5(1)(0-9)

=45