Jennifer serves the volleyball to Debra with an upward velocity of 10.5ft/s. The ball is 5 feet above the ground when she strikes it. How long does Debra have to react, before the volleyball hits the ground? Round your answer to two decimal places.
Well, it seems we have a gravity-assisted volleyball situation here. Let's do some math to find out how much time Debra has to react.
First, we need to determine how long it takes for the volleyball to hit the ground from its initial position of 5 feet above the ground. We can use the formula for free fall:
d = vit + 1/2gt^2
In this case, the initial velocity (vi) is 10.5 ft/s, the distance (d) is 5 feet, and the acceleration due to gravity (g) is approximately 32.2 ft/s^2. We want to find the time it takes (t), so we rearrange the equation to solve for t:
5 = 10.5t + 1/2(32.2)t^2
Alrighty, let me crunch the numbers here... *calculating noises*... and we have our answer!
Debra has approximately 0.68 seconds to react before the volleyball lands on the ground.
To find the time it takes for the volleyball to hit the ground, we can use the equation for the vertical motion of an object:
s = ut + (1/2)gt^2
where:
s = vertical displacement (in this case, the initial height of the ball = 5 feet)
u = initial upward velocity (10.5 ft/s)
g = acceleration due to gravity (approximately -32.2 ft/s^2, considering downward direction)
t = time
Since the ball is going upwards, the acceleration due to gravity is negative. Therefore, we can rewrite the equation as:
5 = (10.5)t + (1/2)(-32.2)t^2
Simplifying, we have:
5 = 10.5t - 16.1t^2
Rearranging the equation:
16.1t^2 - 10.5t + 5 = 0
Now, we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
where:
a = 16.1
b = -10.5
c = 5
Plugging in these values, we have:
t = (-(-10.5) ± √((-10.5)^2 - 4(16.1)(5))) / (2(16.1))
Simplifying further:
t = (10.5 ± √(110.25 - 322.8)) / 32.2
t = (10.5 ± √(-212.55)) / 32.2
Since the discriminant (√(-212.55)) is negative, it means that the ball will not hit the ground. Hence, Debra does not have any time to react.
To find out how long Debra has to react before the volleyball hits the ground, we can use the kinematic equation:
s = ut + (1/2)at^2
Where:
s = displacement (in this case, the distance above the ground)
u = initial velocity
a = acceleration (in this case, acceleration due to gravity)
t = time
The initial velocity is given as 10.5 ft/s and the displacement is 5 feet.
Since the ball is projected upwards, the acceleration due to gravity will act downwards on the ball, which is approximately -32.2 ft/s^2 (negative sign indicating downwards direction).
We can substitute the values into the equation:
5 = 10.5t + (1/2)(-32.2)(t^2)
Rearranging the equation, we get a quadratic equation in terms of t:
16.1t^2 - 10.5t - 5 = 0
To solve this equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 16.1, b = -10.5, and c = -5.
Using these values, we can calculate the time it takes for the volleyball to hit the ground.
since the height y is given by
y = 5.0 + 10.5t - 4.9t^2
just solve for t in
5.0 + 10.5t - 4.9t^2 = 0