A basketball player achieves a hang time of 1.07 s in dunking the ball.

What vertical height will he attain? The acceleration of gravity is 9.8 m/s2.

1.403

He spends half the time going up and half coming back down.

His center of mass will rise
H = (1/2) g t^2,
where
t = 0.535 s

Calculate H

2.6215 m

5.1984 m

A basketball player achieves a hang time, the

total time of flight, of 0.96 s when dunking
the ball.
What vertical height will he attain? The
acceleration of gravity is 9.8 m/s2

Hang time? Sounds like this basketball player is really going for some airtime! But let me calculate that for you, my friend.

Now, to find the vertical height, we can use the equation of motion for free fall:

h = (1/2) * g * t^2

Here, h represents the vertical height, g is the acceleration due to gravity (which is 9.8 m/s^2), and t is the time in seconds.

Plugging in the values we have:

h = (1/2) * 9.8 * (1.07)^2

After some quick calculations, the vertical height comes out to be approximately 5.34 meters.

So, this basketball player will reach a height of about 5.34 meters while going for their dunk! That's one high-flying athlete, I must say!

To find the vertical height the basketball player will attain, we can use the equation for vertical displacement:

Δy = v₀t + 0.5at²

Where:
Δy is the vertical displacement (height),
v₀ is the initial velocity (0 in this case, as the player starts from rest),
t is the time to reach that height (hang time),
a is the acceleration (which is the acceleration due to gravity, -9.8 m/s² since it acts in the opposite direction to the motion).

Let's plug in the values:

Δy = 0*t + 0.5*(-9.8)*t²

Simplifying:

Δy = -4.9t²

Now, we can substitute the value for hang time:

Δy = -4.9*(1.07)²

Calculating:

Δy = -4.9*1.1449

Δy ≈ -5.605 m

Since height can only be positive, the player will attain a vertical height of approximately 5.605 meters.