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

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

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 = (-b\pm√(b^2 - 4ac))/(2a)`

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

`\displaystyle t = (-(-8)\pm√((-8)^2-4(1)(6)))/(2(1))`

Simplify:

`\displaystyle t = (8\pm√(40))/(2) = (8\pm 2√(10))/(2) = 4\pm √(10)`

Hence, our two solutions are:

`\displaystyle t = 4+√(10) \approx 7.16\text{ or } t= 4-√(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.