Projectile Motion cannonball

[Timmy96]

Homework Statement

A 2.5kg cannonball is fired from the top of a 75m cliff at an angle of 30 degrees above the horizontal with an initial speed of 120m/s.

Homework Equations

What is the minimum speed of the cannonball?
When does the cannonball hit the ground?
What is it's speed just before it hits the ground?

The Attempt at a Solution

I don't really have an attempt as I am not sure where to start but I would prefer guidance on how to complete the problem instead of just receiving answers.

 

[SteamKing]

Homework Statement

A 2.5kg cannonball is fired from the top of a 75m cliff at an angle of 30 degrees above the horizontal with an initial speed of 120m/s.

Homework Equations

What is the minimum speed of the cannonball?
When does the cannonball hit the ground?
What is it's speed just before it hits the ground?

The Attempt at a Solution

I don't really have an attempt as I am not sure where to start but I would prefer guidance on how to complete the problem instead of just receiving answers.

What happens to the cannon ball after it is fired; that is, what forces are acting on the ball itself while it is aloft?

 

[Timmy96]

just gravity

 

[BvU]

Good. Now write down some relevant equations under 2. And show what you have done under 3.

 

[HalfLostAndSearching]

I can provide guidance on how to approach this problem using the principles of projectile motion.

First, we need to understand the initial conditions of the cannonball's motion. We know that the cannonball is fired from the top of a 75m cliff at an angle of 30 degrees above the horizontal with an initial speed of 120m/s. This means that the initial velocity of the cannonball can be broken down into two components: a horizontal component and a vertical component. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.

Next, we can use the equations of projectile motion to solve for the minimum speed of the cannonball. The minimum speed occurs at the peak of the cannonball's trajectory, where the vertical velocity is equal to zero. We can use the equation v_f = v_i + at to solve for the minimum speed, where v_f is the final velocity (in this case, zero), v_i is the initial velocity (120m/s), a is the acceleration due to gravity (-9.8m/s^2), and t is the time it takes for the cannonball to reach its peak.

To solve for t, we can use the equation y = y_i + v_iy*t + 1/2at^2, where y is the vertical position, y_i is the initial vertical position (75m), v_iy is the initial vertical velocity (found by breaking down the initial speed of 120m/s into its vertical component using trigonometry), and a and t are the same as before.

Once we have solved for t, we can plug that value back into the equation for the minimum speed to find the answer.

To find when the cannonball hits the ground, we can use the same equation for y, but this time we want to solve for t when y = 0 (since the cannonball will hit the ground when it reaches a vertical position of zero). This will give us the total time it takes for the cannonball to hit the ground.

Finally, to find the speed of the cannonball just before it hits the ground, we can use the equation v_f = v_i + at once again, where v_i is the initial velocity (which we know is 120m/s) and a is the acceleration due to gravity. This will give us the final velocity of the cannonball just before it hits the ground.