Question

height of an object t seconds after it is launched is modeled by the function h(t) = -5t^2 + 30t + 200. the object will hit the ground in how many seconds.

Answer 1

Answer: 10 seconds

Explanation:

When the object hits the ground, the height (y-value) is 0.

h(t) = -5t² + 30t + 200

0 = -5t² + 30t + 200 set height equal to zero

0 = -5(t² - 6t - 40) factored out -5

0 = -5(t - 10)(t + 4) factored the quadratic

0 = t - 10 0 = t + 4 applied the Zero Product Property

t = 10 t = -4 solved for t

↓

t = -4 is not valid (since time cannot be negative)

So, t = 10

Answer 2

10 seconds

Explanation:

AS it is given in the question that

h(t) = -5t² + 30t + 200 ...........................(i)

We have to find the time when it will touch the ground

so it will be touching the ground when its height from ground will be zero

i.e. h(t)=0

so equation (i) becomes

0 = -5t² + 30t + 200

Now we will solve this to get the value of t

Dividing whole equation with (-5) it will give us

t² - 6t - 40 = 0

Using the mid term breaking rule which will break the mid term and we will get the factors from the mid term

t² - 10t + 4t - 40 = 0

Now we will do the factorization of it

t(t-10)+4(t-10)=0

(t-10)(t+4)=0

so either

t-10 = 0 or t+4=0

t=10 or t= - 4

as t is time so it can not be negative

so

t=10 seconds will be the time in which it will touch the earth