Question

A 10 m long vertical cannon is used to accelerate a 1.0 kg ball straight up into air. A constant force of 13.2N is used to access bowling ball length of the canon. what is the ball's approximate velocity as it leaves the canon assuming no energy loss to friction​

Answer 1

Answer:

13.2 n

Explanation:

Answer 2

The ball's approximate velocity as it leaves the cannon is 16.2 m/s, assuming no energy loss to friction.

To calculate the velocity of the ball as it leaves the cannon, we can use the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy.

We can use this relationship to find the initial velocity of the ball as it leaves the cannon.

First, we need to calculate the work done on the ball by the constant force of 13.2 N.

The work done by a constant force is given by the equation:

W = Fd

Where W is the work done, F is the force, and d is the distance traveled by the force.

In this case, the distance traveled is equal to the length of the cannon, or 10 m. So,

W = Fd

$= 13.2 N * 10 m$

= 132 J

ΔKE = W = 132 J

The initial kinetic energy of the ball is zero, since it starts from rest.

So, the final kinetic energy of the ball is equal to 132 J.

We can use this to find the velocity of the ball as it leaves the cannon:

KE$= 0.5 * m * v^2$

$v^2 = 2 * KE / m$

= $2 * 132 J / 1 kg$

= $264 m^2/s^2$

v = √($264 m^2/s^2$) = 16.2 m/s

So, the ball's approximate velocity as it leaves the cannon is 16.2 m/s, assuming no energy loss to friction.

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