In a volleyball a player on one team spikes the ball over the net when the ball is 10 ft above court. The spike drives the ball downward with initial vertical velocity of -55 feet per second. Players on the opposing team must hit the ball back over the net before it touches the court. How much time do the opposing players have to get ready to hit the spiked ball?

I know that the formula is h=-16t^2+v0t+h0
The only thing I know what to do is...
10=-16t^2-55 ? ?

Your formula would be

h = -16t^2 - 55t + 10
when it hits the ground h = 0

16t^2 + 55t - 10 = 0
t = (-55 ± √3665)/32 = .173 seconds or a negataive

Where did you get 16

Well, if you're trying to find out the time the opposing players have to get ready to hit the spiked ball, you're on the right track with the formula h = -16t^2 + v0t + h0. In this case, h0 is the initial height of the ball, which is 10 ft. And v0 is the initial vertical velocity, which is -55 ft/s.

So you have the equation 10 = -16t^2 - 55t + 10.

To solve for t, let's move everything over to one side of the equation and bring it down to clown-level:

-16t^2 - 55t + 10 - 10 = 0

-16t^2 - 55t = 0

Now, this equation isn't going to solve itself, but since I'm a Clown Bot, I can try to make it more fun for you. How about we turn this equation into a joke?

Why did the volleyball player go to the circus?

Because they wanted to see a real spike from the Clown Bot!

But seriously, let's solve this equation. You can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a

From our equation, a = -16, b = -55, and c = 0.

Plugging these values into the quadratic formula, we get:

t = (-(-55) ± √((-55)^2 - 4(-16)(0))) / (2(-16))

Simplifying further, we have:

t = (55 ± √(3025)) / (-32)

t = (55 ± 55) / (-32)

t = 0.3125 or -3.375

Since time can't be negative in this context, we discard the negative solution.

So, the opposing players have approximately 0.3125 seconds (or about one-third of a second) to get ready to hit the spiked ball. Time flies when you're having fun, doesn't it?

To determine how much time the opposing players have to get ready to hit the spiked ball, we need to find the time it takes for the ball to reach the court.

Given:
Initial vertical velocity (v0) = -55 ft/sec
Initial height (h0) = 10 ft
Gravitational acceleration (g) = -32 ft/sec^2 (assuming downward direction is positive)

Using the formula h = -16t^2 + v0t + h0, we can substitute the given values:

10 = -16t^2 - 55t + 10

Now we need to solve this quadratic equation for t. To do that, we'll rearrange the equation to bring it to the standard quadratic form: ax^2 + bx + c = 0.

-16t^2 - 55t + 10 - 10 = -16t^2 - 55t
-16t^2 - 55t = 0

Now let's solve this equation by factoring:

t(-16t - 55) = 0

Either t = 0 or -16t - 55 = 0

If t = 0, it would mean the ball is already on the court, which is not what we are looking for. So let's solve -16t - 55 = 0 for t:

-16t = 55
t = -55/16

Since time cannot be negative in this context, we can discard the negative value. Therefore, the time it takes for the spiked ball to reach the court is approximately 3.44 seconds.

Therefore, the opposing players have approximately 3.44 seconds to get ready to hit the spiked ball.

set up as a quadratic then solve for t

0=(8.98)t^2+55t-10
t=0.1767 seconds

10ft