Calculate the hang time of a person who moves horizontally 2.00 m during a 1.23 m high jump.
Express your answer with the appropriate units
The motion of jumping person is similar to projectile motion:
H=(Vo)^2•(sinα)^2/2g
L=(Vo)^2•(sin2α)/2g=
=(Vo)^2•(2sinα•cosα)/2g.
2H/L=sinα/cosα,
tanα=2H/L=21.23/2=1.23
α=arctan1.23=50.9°
Vo=sqr(2gH)/ sinα=6.33 m/s.
t=2Vo•sinα/g=1 s.
To calculate the hang time of a person during a jump, we need to find the time it takes for the person to reach the maximum height of the jump first.
To find the time it takes for the person to reach the maximum height, we can use the vertical motion equation:
y = v0y * t + 0.5 * a * t^2
where:
y = vertical displacement (1.23 m)
v0y = initial vertical velocity (we assume it to be 0 m/s, as the person starts from rest)
a = acceleration due to gravity (-9.8 m/s^2)
t = time it takes to reach the maximum height
Plugging in the values, we have:
1.23 m = 0 * t + 0.5 * (-9.8 m/s^2) * t^2
Simplifying the equation, we get:
4.9 t^2 = 1.23
Dividing both sides by 4.9, we have:
t^2 = 1.23 / 4.9
t^2 = 0.25
Taking the square root of both sides, we find:
t = √0.25
t ≈ 0.5 s
Therefore, it takes approximately 0.5 seconds for the person to reach the maximum height of the jump.
Now that we know the time it takes for the person to reach the maximum height, we can calculate the total hang time by doubling the time. This is because the time taken to reach the maximum height is the same as the time taken to fall back down and land. Therefore:
Total hang time = 2 * t
Total hang time = 2 * 0.5 s
Total hang time ≈ 1.0 s
Thus, the hang time of the person during the jump is approximately 1.0 second.