Hang time?
If a basketball player has a vertical leap of about 30 inches, what is his hang time? V-48T^2
Hang time to the nearest tenth in seconds.
:-)
To calculate the hang time of the basketball player, we can use the hang time formula: V = 48 * T^2, where V represents the vertical leap in inches and T represents the hang time in seconds.
In this case, the vertical leap is given as 30 inches. We can substitute this value into the formula:
30 = 48 * T^2
Now, let's solve for T. Divide both sides of the equation by 48:
30/48 = T^2
0.625 = T^2
To find T, we need to take the square root of both sides of the equation. However, since the measurement is for time, we discard the negative solution:
√0.625 = T
T ≈ ±0.791
Since the value of time can't be negative, we discard the negative solution. Therefore, the approximate hang time is 0.791 seconds, rounded to the nearest tenth.
To find the hang time, we can use the formula V = 48T^2, where V represents the vertical leap in inches and T is the hang time in seconds.
Given that the basketball player has a vertical leap of about 30 inches, we can substitute V = 30 into the formula:
30 = 48T^2
To find T, we need to isolate the T^2 term. Divide both sides of the equation by 48:
30/48 = T^2
Simplify the fraction:
5/8 = T^2
To find T, take the square root of both sides:
√(5/8) = √(T^2)
Simplify:
√(5/8) = T
To calculate the square root of 5/8, we get approximately 0.707. Therefore, the hang time T is approximately 0.707 seconds, rounded to the nearest tenth.
So the basketball player's hang time is about 0.7 seconds.