Question

A fireworks display is launched from a platform 10 feet above ground with an initial upward velocity of 70 feet per second. The height of the fireworks above ground after t seconds is given by the equation h = –16t2 + 70t + 10, where h is the height of the fireworks in feet and t is the time in seconds after they are launched. What is the maximum height of the fireworks display, to the nearest foot?
A.
87 feet
B.
174 feet
C.
70 feet
D.
10 feet

Answer 1

Hello there.

A fireworks display is launched from a platform 10 feet above ground with an initial upward velocity of 70 feet per second. The height of the fireworks above ground after t seconds is given by the equation h = –16t2 + 70t + 10, where h is the height of the fireworks in feet and t is the time in seconds after they are launched. What is the maximum height of the fireworks display, to the nearest foot?

C.
70 feet

Answer 2

Answer:

87 ft

Step-by-step explanation:

We are given that the height of the fireworks above the ground after t seconds is given by

$h(t)=-16t^2+70t+10$

Initial velocity of firework display=70 m/s

We have to find the maximum height of the fireworks display .

Differentiate w.r.t t

$h'(t)=-32t+70$

Substitute h'(t)=0

$-32t+70=0$

$-32t=-70$

$t=\frac{70}{32}=\frac{35}{16}$

Again differentiate w.r.t t

$h''(t)=-32 <0$

Hence, height is maximum at t=$\frac{35}{16}$

Substitute $t=\frac{35}{16}$

Then, we get

$h(\frac{35}{16})=-16(\frac{35}{16})^2+70\times \frac{35}{16}+10=87 ft$

Hence, the maximum height of the fireworks display=87 ft

Answer: 87 ft