An object that is falling or vertically projected into the air has it height, in feet, above the ground given by the formula below, where s is the height in feet, Vo is the original velocity of the object, in feet per second, t is the time the object is in motion in seconds, and So is the height, in feet, from which the object is dropped or projected. The figure shows that a ball is thrown straight up from a rooftop at an original velocity of 24 feet per second from a height of 96 feet. The ball misses the rooftop on its way down. How high above the ground will the ball be 2 seconds after thrown?

s=16t^2+VottSo. Please help, I'm stuck

fix the typo, and plug in the numbers

s = 96 + 24t - 16t^2
when t=2,
s(2) = 96+48-64 = 80

Oh, don't worry, I'm here to help!

To find the height of the ball 2 seconds after it is thrown, we can use the given formula:

s = 16t^2 + Vot + So

Let's plug in the values given in the problem:

t = 2 (the time in seconds)
Vo = 24 (the original velocity in feet per second)
So = 96 (the height from which the object is projected)

Now we can substitute those values into the formula:

s = 16(2)^2 + 24(2) + 96

And let's simplify it:

s = 16(4) + 48 + 96
s = 64 + 48 + 96
s = 208

So, the ball will be 208 feet above the ground 2 seconds after it is thrown.

Keep in mind, though, that the ball in the problem misses the rooftop on its way down, so the final answer is not relative to the rooftop. It's just the height above the ground.

Hope that helps! Let me know if you have any more questions.

To find the height of the ball 2 seconds after it is thrown, we can use the formula:

s = 16t^2 + Vot + So

Given:
Vo (original velocity) = 24 feet per second
So (initial height) = 96 feet
t (time) = 2 seconds

Substituting the values into the formula:

s = 16(2)^2 + (24)(2) + 96

Calculating the values within parentheses:

s = 16(4) + 48 + 96

Multiplying 16 by 4:

s = 64 + 48 + 96

Adding the values together:

s = 208

Therefore, the ball will be 208 feet above the ground at 2 seconds after being thrown.

To find the height of the ball 2 seconds after it is thrown, we can use the given formula:

s = 16t^2 + Vo*t + So

Given:
Vo = 24 feet per second (original velocity)
So = 96 feet (initial height)

We need to substitute these values into the formula and calculate the height at t = 2 seconds.

s = 16(2)^2 + 24(2) + 96

First, let's calculate 16(2)^2:
16(2)^2 = 16(4) = 64

Next, let's calculate 24(2):
24(2) = 48

Now we can substitute these values into the formula:

s = 64 + 48 + 96

Let's simplify the equation:

s = 208

Therefore, the ball will be 208 feet above the ground 2 seconds after it is thrown.