Question

The height of a projectile fired upward is given by the formula
s = v0t − 16t2,
where s is the height in feet,
v0
is the initial velocity, and t is the time in seconds. Find the time for a projectile to reach a height of 96 ft if it has an initial velocity of 128 ft/s. Round to the nearest hundredth of a second.

Answer 1

Answer:

The projectile will reach a height of 96 feet after about 0.84 seconds as well as after about 7.16 seconds.

Step-by-step explanation:

The height of a projectile fired upward is given by the formula:

$\displaystyle s = v_{0} t - 16t^2$

Where s is the height in feet, v₀ is the initial velocity, and t is the time in seconds.

Given a projectile with an initial velocity of 128 ft/s, we want to determine how long it will take the projectile to reach a height of 96 feet.

In other words, given that v₀ = 128, find t such that s = 96.

Substitute:

$(96) = (128)t-16t^2$

This is a quadratic. First, we can divide both sides by -16:

$-6 = -8t+t^2$

Isolate the equation:

$t^2 - 8t + 6 = 0$

The equation isn't factorable, so we can consider using the quadratic formula:

$\displaystyle t = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$

In this case, a = 1, b = -8, and c = 6. Substitute:

$\displaystyle t = \frac{-(-8)\pm\sqrt{(-8)^2-4(1)(6)}}{2(1)}$

Simplify:

$\displaystyle t = \frac{8\pm\sqrt{40}}{2} = \frac{8\pm 2\sqrt{10}}{2} = 4\pm \sqrt{10}$

Hence, our two solutions are:

$\displaystyle t = 4+\sqrt{10} \approx 7.16\text{ or } t= 4-\sqrt{10} \approx 0.84$

So, the projectile will reach a height of 96 feet after about 0.84 seconds as well as after about 7.16 seconds.