Question

A projectile is fired upward from a height of 160 feet above the ground, with an initial velocity of 600ft/sec. Recall that projectiles are modeled by the function h(t)=−16t2+v0t+y0. How long will the projectile be in flight? Round your answer to the nearest hundredth.

Answer 1

t ≈ 37.76 seconds (rounded to the nearest hundredth)

Explanation:

To solve the problem, we can use the equation for the height of a projectile at time t: h(t) = - 16 * t^2 + v0 * t + y0

where v0 is the initial velocity and y0 is the initial height.

In this case, we have: v0 = 600 ft/sec (upward)

y0 = 160 ft (above the ground)

We want to find the time at which the projectile hits the ground, which is when h(t) = 0.

So we can set up the equation: 0 = -16 * t^2 + 600 * t + 160

We can solve for t using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = -16, b = 600, and c = 160.

Plugging in the values, we get:

t = (-600 ± sqrt(600^2 - 4(-16)(160))) / 2(-16)

t = (-600 ± sqrt(360000 + 10240)) / (-32)

t = (-600 ± sqrt(370240)) / (-32)

We can simplify this expression by taking the negative root since the positive root would give a negative time, which is not physically meaningful in this context: t = (-600 - 608.47) / (-32)

t ≈ 37.76 seconds (rounded to the nearest hundredth)

Therefore, the projectile will be in flight for about 37.76 seconds before hitting the ground.