A ball rolls horizontally off the edge of a tabletop that is 1.60 m high. It strikes the floor at a point 1.58 m horizontally away from the table edge. (Neglect air resistance.)

(a) How long was the ball in the air?
___ s
(b) What was its speed at the instant it left the table?
___ m/s

.. am i trying to find the angle?

theta = tan^-1(x/h) ?

To find the answers to the given questions, we can use the equations of motion and the principle of conservation of energy. Let's break it down step by step:

(a) How long was the ball in the air?

We can solve this problem using kinematic equations.
The vertical distance traveled by the ball is the height of the tabletop, which is 1.6 m.
The horizontal distance traveled by the ball is 1.58 m.

First, we need to find the time it takes for the ball to reach the ground.

Using the equation of motion: s = ut + (1/2)at²
Where s = vertical distance (1.6 m), u = initial vertical velocity (0 m/s), a = acceleration due to gravity (-9.8 m/s²), and t = time.

Plugging in the values, we get: 1.6 = 0*t + (1/2)(-9.8)t²
This simplifies to: 4.9t² = 1.6
Solving for t, we find: t = sqrt(1.6/4.9) ≈ 0.8 seconds.

Therefore, the ball was in the air for approximately 0.8 seconds.

(b) What was its speed at the instant it left the table?

We can use the principle of conservation of energy to find the initial speed.

The initial potential energy (PE) of the ball when it is on the tabletop converts into kinetic energy (KE) when it reaches the ground.

The potential energy at the tabletop is given by: PE = mgh
Where m = mass of the ball, g = acceleration due to gravity (9.8 m/s²), and h = height of the table (1.6 m).

The kinetic energy of the ball at the instant it leaves the table is given by: KE = (1/2)mv²
Where v is the velocity of the ball.

Since energy is conserved, we have: PE = KE
Therefore, mgh = (1/2)mv²

Canceling out the mass on both sides, we have: gh = (1/2)v²
Rearranging the equation, we get: v = √(2gh)

Plugging in the values, v = √(2*9.8*1.6) ≈ 6.26 m/s.

Therefore, the speed of the ball at the instant it left the table was approximately 6.26 m/s.

In summary:
(a) The ball was in the air for approximately 0.8 seconds.
(b) The speed of the ball at the instant it left the table was approximately 6.26 m/s.

No, in this case, we don't need to find the angle.