Question

A projectile is thrown upward so that its distance above the ground after t seconds is h =-13t2 + 390t. After how many seconds does it reach its maximum height? A) 22.5 s B) 7 s C) 15 s D) 30 s

Answer 1

Answer:

The option C) 15 s is correct

Therefore the maximum height it reaches in 15 seconds.

Step-by-step explanation:

Given equation is $h =-13t^2 + 390t$

Given that a projectile is thrown upward so that its distance above the ground after t seconds is $h =-13t^2 + 390t$

To find how many seconds does it reach its maximum height:

"The standard form of a parabola's equation is  expressed as :

$y=ax^2+bx+c$.

If $a>0$, then the parabola opens upwards;

if $a$ the parabola opens downwards."

The maximum height is the vertex of the parabola  $h =-13t^2 + 390t$ which is $\frac{-b}{2a}$.

Comparing the given equation with the standard form of parabola we get

the values of a=-13 , b=390 and c=0

The maximum height in $t=\frac{-b}{2a}$.

Substitute the values we get

$t=-\frac{390}{2(-13)}$

$=\frac{390}{26}$

$=15$

∴ t=15 s

∴ The option C) 15 s is correct.

∴ the maximum height is the vertex of the parabola, at t reaches in 15 seconds.