Question

the equation that models the height of an aerial bomb (t) seconds after it is shot upwards from the ground with an initial velocity of ft/sec is given by (h=-16t^{2}+80t). the maximum height above ground level that the aerial bomb will reach is 100 feet. how many seconds after it was launched will it take to reach that height?
a 2.5 seconds
b 3 seconds
c 2 seconds
d 4 seconds

Explanation:

Step1: Set height equation to maximum height

Set $h = 100$ in $h=-16t^{2}+80t$, getting $100=-16t^{2}+80t$.

Step2: Rearrange to standard quadratic form

Rearrange to $16t^{2}-80t + 100 = 0$. Divide by 4: $4t^{2}-20t + 25 = 0$.

Step3: Use quadratic formula or factor

Factor the quadratic: $(2t - 5)^{2}=0$.

Step4: Solve for $t$

Set $2t-5 = 0$, then $2t=5$, so $t = 2.5$.

Answer:

A. 2.5 seconds