The professor throws a ball straight up. The ball leave's the prof's hand 2.0 m above the ground, with a velocity of 15 m/s. How long is the ball in the air before it hits the ground? If the professor catches the ball at the same height from which it is release, what is its velocity when caught?

The ball is 3.2 s in the air before it hits the ground.

When caught, the velocity will be - 15 m/s.

Explanation:

The equations for the position and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

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

v = velocity at time "t".

Let's place the origin of the frame of reference on the ground so that when the ball hits the ground its height is 0. Then, using the equation of height, we can calculate the time at which the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 2.0 m + 15 m/s · t - 1/2 · 9.8 m/s² · t²

0 m = 2.0 m + 15 m/s · t - 4.9 m/s² · t²

Solving the quadratic equation:

t = 3.2 s (the other solution of the equation is negative, therefore, it was discarded)

The ball is 3.2 s in the air before it hits the ground.

Now, let's calculate how much time it takes the ball to return to a height of 2.0 m.

y = y0 + v0 · t + 1/2 · g · t²

2.0 m = 2.0 m + 15 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (15 m/s - 1/2 · 9.8 m/s² · t) t = 0 (initial position)

0 = 15 m/s - 1/2 · 9.8 m/s² · t

-15 m/s / - 4.9 m/s² = t

t = 3.1 s

Now, using the equation of velocity, we can calculate the velocity at t = 3.1 s when the ball is caught. Notice that since the acceleration is constant, the velocity should be equal to the initial velocity but in opposite direction. Instead of 3.1 s, I will use t = - 15 m/s / - 4.9 m/s² to avoid error due to rounding.

v = v0 + g · t

v = 15 m/s - 9.8 m/s² · (-15 m/s / - 4.9 m/s²)

v = - 15 m/s

When caught, the velocity will be - 15 m/s