You are playing volleyball. You hit the volleyball with an initial vertical velocity of 15 feet per second when it is 4 feet above the ground. Find how many seconds it takes to reach the maximum height, the maximum height reached, and the time the object is in the air.
a. V = Vo + g*t
Tr = (V-Vo)/g = (0-15)/-32 = 0.469 s. To
reach max. Ht. = Rise time.
b. h = ho + Vo*t + 0.5g*t^2
h = 4 + 15*0.469 - 16*(0.469)^2=7.52Ft.
Above Gnd.
c. h = Vo*t + 0.5g*t^2 = 7.52 Ft.
0 + 16*T^2 = 7.52
T^2 = 0.469
Tf = 0.685 s. = Fall time.
T = Tr + Tf = 0.469 + 0.685 = 1.15 s. in
air.
To find the time it takes to reach the maximum height, we can use the formula for vertical motion:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time
Since the volleyball reaches its maximum height, the final velocity at the maximum height is 0 ft/s (the ball momentarily stops before falling back down). The initial velocity (vi) is 15 ft/s (given) and the acceleration (a) is equal to the acceleration due to gravity, which is approximately -32 ft/s² (assuming upward motion as positive and downward motion as negative).
0 = 15 + (-32)t
Rearranging the equation, we have:
32t = 15
t = 15 / 32 ≈ 0.46875 seconds
So, it takes approximately 0.46875 seconds to reach the maximum height.
To find the maximum height reached, we can use the formula for vertical motion:
h = vi * t + 0.5 * a * t²
Where:
h = vertical displacement (in this case, the maximum height)
vi = initial velocity
a = acceleration
t = time
Substituting the values:
h = 15 * 0.46875 + 0.5 * (-32) * (0.46875)²
h = 7.03125 - 0.703125
h ≈ 6.32813 feet
So, the maximum height reached is approximately 6.32813 feet.
To find the time the object is in the air, we can double the time it takes to reach the maximum height since the object will be in the air both on the way up and on the way down.
Time in the air = 2 * 0.46875 ≈ 0.9375 seconds
Therefore, the time the object is in the air is approximately 0.9375 seconds.
To find the time it takes to reach the maximum height, we can use the fact that the initial vertical velocity and the acceleration due to gravity will cause the ball to eventually reach a vertical velocity of 0 at the maximum height.
The equation that relates vertical velocity, time, and acceleration due to gravity is given by the formula:
vf = vi + gt,
where vf is the final vertical velocity, vi is the initial vertical velocity, g is the acceleration due to gravity (approximately -32.2 feet per second squared), and t is the time.
In this case, we know that the final vertical velocity is 0, the initial vertical velocity is 15 feet per second, and the acceleration due to gravity is -32.2 feet per second squared.
So, we can substitute these values into the equation and solve for t:
0 = 15 + (-32.2)t.
Rearranging the equation, we have:
32.2t = 15.
Dividing both sides by 32.2, we find:
t ≈ 0.465 seconds.
Therefore, it takes approximately 0.465 seconds to reach the maximum height.
To find the maximum height reached, we can use the fact that the final vertical velocity at that point is 0.
The equation that relates vertical displacement, initial vertical velocity, time, and acceleration due to gravity is given by the formula:
hf = hi + vit + 0.5gt^2,
where hf is the final vertical displacement (maximum height), hi is the initial vertical displacement (4 feet), vi is the initial vertical velocity (15 feet per second), g is the acceleration due to gravity (approximately -32.2 feet per second squared), and t is the time.
In this case, we know that the final vertical displacement is the maximum height, the initial vertical displacement is 4 feet, the initial vertical velocity is 15 feet per second, and the acceleration due to gravity is -32.2 feet per second squared.
Substituting these values into the equation, we have:
hf = 4 + 15t + 0.5(-32.2)t^2.
Setting hf to 0 (since the final vertical velocity at the maximum height is 0), we get:
0 = 4 + 15t + 0.5(-32.2)t^2.
Simplifying the equation, we have:
0 = 0.5(-32.2)t^2 + 15t + 4.
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a,
where a, b, and c are the coefficients of the quadratic equation. In this case, a = 0.5(-32.2), b = 15, and c = 4.
Evaluating the equation, we find:
t ≈ 0.774 seconds,
or
t ≈ 0.097 seconds.
Since the ball reaches the maximum height after 0.465 seconds, we discard the solution t ≈ 0.097 seconds (which corresponds to the time it takes for the ball to reach the highest point during its descent).
Therefore, the ball reaches the maximum height after approximately 0.465 seconds.
To find the total time the object is in the air, we double the time it takes to reach the maximum height:
Total time = 2 × 0.465 seconds = 0.93 seconds.
Therefore, the ball is in the air for approximately 0.93 seconds.