Debra serves the volleyball to Grace with an upward velocity of 24.5ft/s.
The ball is 5 feet above the ground when she strikes it. How long does Grace have to react, before the volleyball hits the ground? Round your answer to two decimal places.
Well, it seems like Debra and Grace have quite the game going on! Let's calculate how much time Grace has to react.
First, we need to find out how long it takes for the volleyball to hit the ground. We can use the equation of motion:
s = ut + (1/2)at^2
Here, s is the displacement (5 feet), u is the initial velocity (24.5 ft/s), a is the acceleration due to gravity (-32.2 ft/s^2), and t is the time.
By rearranging the equation, we get:
t = sqrt((2s)/a)
Substituting the values, we have:
t = sqrt((2 * 5) / -32.2)
t ≈ sqrt(0.15) ≈ 0.39 seconds
So, round to two decimal places, Grace has approximately 0.39 seconds to react. Better get those reflexes ready, Grace!
To find the time it takes for the volleyball to hit the ground, we can use the formula for vertical motion:
h = h0 + v0*t - (1/2)gt^2
In this case, h0 is the initial height of the ball (5 feet), v0 is the upward velocity (24.5 ft/s), g is the acceleration due to gravity (-32.2 ft/s^2), and h is the height of the ball when it hits the ground (0 feet).
Plugging in these values, we get:
0 = 5 + 24.5t - (1/2)(-32.2)t^2
Rearranging the equation, we get a quadratic equation:
0 = -0.5(32.2)t^2 + 24.5t + 5
We can solve this equation to find the time it takes for the ball to hit the ground. Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -0.5(32.2), b = 24.5, and c = 5. Plugging in these values, we get:
t = (-24.5 ± √(24.5^2 - 4(-0.5(32.2))(5))) / (2(-0.5(32.2)))
Simplifying further:
t = (-24.5 ± √(600.25 - (-322)))) / (-32.2)
t = (-24.5 ± √(600.25 + 322)) / -32.2
t = (-24.5 ± √922.25) / -32.2
Taking the positive root:
t = (-24.5 + √922.25) / -32.2
t ≈ (-24.5 + 30.36) / -32.2
t ≈ 5.86 / -32.2
t ≈ -0.182
Since time cannot be negative in this context, we can ignore the negative solution.
Therefore, Grace has approximately 0.18 seconds to react before the volleyball hits the ground.
To solve this problem, we can use the kinematic equation:
h = h0 + v0t - 0.5gt^2
Where:
- h: final height (0 ft, since the ball hits the ground)
- h0: initial height (5 ft above the ground)
- v0: initial velocity (24.5 ft/s, positive because it's upward)
- g: acceleration due to gravity (32.2 ft/s^2, negative because it's downward)
- t: time (which we need to find)
Plugging in the known values, we have:
0 = 5 + (24.5)t - 0.5(32.2)t^2
Simplifying the equation, we get a quadratic equation:
-16.1t^2 + 24.5t + 5 = 0
To solve for t, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = -16.1, b = 24.5, and c = 5. Plugging in these values, we have:
t = (-24.5 ± sqrt(24.5^2 - 4(-16.1)(5))) / (2(-16.1))
Calculating this expression gives us two possible solutions for t. However, since we are interested in how long Grace has to react before the ball hits the ground, we only consider the positive value for t. So, we have:
t = (-24.5 + sqrt(24.5^2 - 4(-16.1)(5))) / (2(-16.1))
Evaluating this expression gives us the value of t. Rounding it to two decimal places, we get the time Grace has to react before the volleyball hits the ground.