Solve the problem.
-A projectile is thrown upward so that its distance above the ground after $t$ seconds is $h = - 15 t ^ { 2 } + 540 t$. What is its maximum height?
A) $7290 \mathrm { ft }$
B) $4860 \mathrm { ft }$
C) $2430 \mathrm { ft }$
D) $3645 \mathrm { ft }$

Question

Quizplus · Accepted Answer

The maximum height of a projectile in the form $h = -at^2 + bt + c$ can be found using the vertex formula for a parabola, $h_{max} = c + \frac{b^2}{4a}$. Here, $a = 15$, $b = 540$, and $c = 0$. Plugging these values into the formula gives $h_{max} = 0 + \frac{540^2}{4*15} = \frac{291600}{60} = 4860$ feet.

Solve the Problem ttt Seconds Is h=−15t2+540th = - 15 t ^ { 2 } + 540 th=−15t2+540t

Solve the Problem $t$ Seconds Is $h = - 15 t ^ { 2 } + 540 t$