You throw a basketball straight down from the roof of a building. The basketball leaves your hand at time t = 0 moving at 5.00 m/s. Air resistance can be neglected. Find the time when the ball is 5.00 m below where it left your hand.

After 1.64 s, the ball is 5.00 m below the throwing point.

Explanation:

Hi there!

The equation of height of the ball is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference on the ground. We have to find the time at which the height of the ball is 5.00 m below the initial height. Mathematically, we have to find the value of t at which h = h0 - 5.00 m.

Using the equation of height:

h = h0 + v0 · t + 1/2 · g · t²    (h = h0 - 5.00 m)

h0 - 5.00 m = h0 + 5.00 m/s · t + 1/2 · (-9.8 m/s²) · t²

0 = 5.00 m + 5.00 m/s · t - 4.9 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.64 s  (the other solution of the equation is rejected because it is negative).

After 1.64 s, the ball is 5.00 m below the throwing point.

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.