A projectile which is fired from the ground with a velocity of 960 ft/sec, at an angle of 30° above the ground, has a height at time t of h(t)=480t–16t2. Find the time when it hits the ground again.

its 5

height has to equal 0 because its hitting the ground

0=480t-16t^2 factor
0=16t(30-t)
so t=0 or 30
0 is when its fired so it must be 30
so it hits the ground again after 30 seconds

To find the time when the projectile hits the ground, we need to determine when its height is zero.

Given the equation for the height of the projectile at time t: h(t) = 480t - 16t^2

When the projectile hits the ground, its height is zero. So, we need to solve the equation:

0 = 480t - 16t^2

To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula here.

The quadratic formula is given by: t = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = -16, b = 480, and c = 0.

Substituting these values into the quadratic formula, we get:

t = (-(480) ± √((480)^2 - 4(-16)(0))) / (2(-16))
t = (-480 ± √(230400 - 0) / (-32)
t = (-480 ± √230400) / (-32)
t = (-480 ± 480) / (-32)

Simplifying further, we have:

t = (-480 + 480) / (-32) = 0 / (-32) = 0
t = (-480 - 480) / (-32) = -960 / (-32) = 30

Since time cannot be negative in this context, we discard the negative value.

Therefore, the time when the projectile hits the ground again is t = 30 seconds.

To find the time when the projectile hits the ground, we need to determine when the height of the projectile is equal to zero.

The given equation for the height of the projectile is h(t) = 480t - 16t^2, where t represents time in seconds.

To find the time when the projectile hits the ground, we can set the height equation h(t) to zero and solve for t:

0 = 480t - 16t^2

Rearranging the equation, we get:

16t^2 = 480t

Dividing both sides by 16t, we get:

t = 480t / 16

Simplifying further:

t = 30t

Dividing both sides by t (assuming t is not zero), we get:

1 = 30

This is not a true statement, which means our assumption that t is not zero is incorrect.

However, t = 0 represents the initial time when the projectile is fired from the ground. Therefore, the time t = 0 is not the time when the projectile hits the ground again.

To find the correct time when the projectile hits the ground again, we need to solve the equation 0 = 480t - 16t^2 for t. Since this is a quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -16, b = 480, and c = 0. Substituting these values into the quadratic formula, we get:

t = (-(480) ± √((480)^2 - 4(-16)(0))) / 2(-16)

Simplifying further:

t = (-480 ± √(230,400)) / (-32)

Using the square root, we find:

t = (-480 ± 480) / (-32)

This gives us two possible solutions:

1. t = (-480 + 480) / (-32) = 0 / (-32) = 0
2. t = (-480 - 480) / (-32) = (-960) / (-32) = 30

Therefore, the projectile hits the ground again at time t = 0 seconds (initial time) and t = 30 seconds.