Question

A firework is launched upward from the ground at an initial velocity of 60 feet per second.
h = -16t^2 + 60t
How long will it take the fireworks to reach its maximum height?

What is the maximum height that the firework will reach?

When will the fireworks hit the ground?

Answer 1

Answer:

It will take the fireworks $1\dfrac{7}{8}$ seconds to reach its maximum height.

The maximum height that the firework will reach is $56\dfrac{1}{4}$ feet.

The fireworks will hit the ground in $3\dfrac{3}{4}$ seconds.

Step-by-step explanation:

A firework is launched upward from the ground at an initial velocity of 60 feet per second. The function describing this situation is

$h = -16t^2 + 60t$

1. Find t-coordinate of the vertex of the parabola:

$t_v\\ \\=\dfrac{-b}{2a}\\ \\=\dfrac{-60}{2\cdot (-16)}\\ \\=\dfrac{15}{8}\ seconds$

It will take the fireworks $\dfrac{15}{8}=1\dfrac{7}{8}$ seconds to reach its maximum height.

2. Find h-coordinate of the vertex of the parabola:

$h(t_v)\\ \\=-16\cdot \left(\dfrac{15}{8}\right)^2+60\cdot \left(\dfrac{15}{8}\right)\ \\=-16\cdot \dfrac{225}{64}+\dfrac{30\cdot 15}{4}\\ \\=\dfrac{-225+450}{4}\\ \\=\dfrac{225}{4}\\ \\=56\dfrac{1}{4}\ ft$

The maximum height that the firework will reach is $56\dfrac{1}{4}$ feet.

3. The fireworks will hit the ground when h = 0:

$-16t^2+60t=0\\ \\4t(-4t+15)=0\\ \\t=0\ \text{or}\ -4t+15=0\\ \\t=0\ \text{or}\ t=\dfrac{15}{4}\ seconds$

The fireworks will hit the ground in $\dfrac{15}{4}=3\dfrac{3}{4}$ seconds.