How do I find final velocity using conservation of energy

[Alameen Damer]

Homework Statement

A paintball gun launches a paintball off a cliff at an angle of elevation of 45°. The cliff is 165 m high. The paintball is initially moving at 180 m/s. Calculate the speed of the paintball as it hits the ground. I'm having trouble understanding how to solve this, I have posted my attempt.

Homework Equations

Ek=1/2mv^2
Eg=mgh

The Attempt at a Solution

I'm going to find the kinetic energy using the initial velocity.

So, cos 45 x 180 = 127.3 m/s. <---- This is velocity in the x direction

Ek=1/2 m *(127.3)^2
=8102.6m

Find height by subbing this into the Eg formula:

8102.6m=mgh
8102.6=9.8h
826.8=h

So this is the height of the projectile, in reference to the cliff. The total height however should have the 165m of the cliff added: Therefore it is (826.8+165)=991.8 m

This is the maximum height, I will use it to find the total energy at max height, so
Eg=991.8gm
=991.8(9.8)m
=9719m

Now I sub this into Ek to find the Velocity.

Ek=9719m
1/2mv^2=9719m
1/2v^2=9719
v^2=19439
v=139 m/s

The book answer says 190 m/s.

 

[Phynos]

I'm going to find the kinetic energy using the initial velocity.

So, cos 45 x 180 = 127.3 m/s. <---- This is velocity in the x direction

Using the x component of velocity to find a height? (It doesn't make a difference mathematically since the angle is 45 degrees, however typically we the x-axis is horizontal) I would have written this off as a typo but you used cosine as well.

This is the maximum height, I will use it to find the total energy at max height, so
Eg=991.8gm
=991.8(9.8)m
=9719m

The total energy of the system should not change, correct? Check what the total energy was at t=0 and compare it with this.

 

[Alameen Damer]

I used cosine because I was trying to find height at the maximum point, where no forces act along the x direction. How can i find total energy at t=0, do i consider it as potential, (before launch)?

 

[Phynos]

I used cosine because I was trying to find height at the maximum point, where no forces act along the x direction.

CosӨ gives the relationship between the x component of the velocity and the original velocity.

EDIT: I should specify that this is only the case when the angle is between the horizontal and the hypotenuse, which is the case in this question.

Heights correspond to the y-axis unless you've switched them for some reason. But since you didn't explicitly state it I assume your going with the convention that X is horizontal and Y is vertical.

How can i find total energy at t, do i consider it as potential, (before launch)?

Add kinetic and potential energy using the initial conditions.

 

[Kilgour22]

Homework Statement

A paintball gun launches a paintball off a cliff at an angle of elevation of 45°. The cliff is 165 m high. The paintball is initially moving at 180 m/s. Calculate the speed of the paintball as it hits the ground. I'm having trouble understanding how to solve this, I have posted my attempt.

Homework Equations

Ek=1/2mv^2
Eg=mgh

The Attempt at a Solution

I'm going to find the kinetic energy using the initial velocity.

So, cos 45 x 180 = 127.3 m/s. <---- This is velocity in the x direction

Ek=1/2 m *(127.3)^2
=8102.6m

Find height by subbing this into the Eg formula:

8102.6m=mgh
8102.6=9.8h
826.8=h

So this is the height of the projectile, in reference to the cliff. The total height however should have the 165m of the cliff added: Therefore it is (826.8+165)=991.8 m

This is the maximum height, I will use it to find the total energy at max height, so
Eg=991.8gm
=991.8(9.8)m
=9719m

Now I sub this into Ek to find the Velocity.

Ek=9719m
1/2mv^2=9719m
1/2v^2=9719
v^2=19439
v=139 m/s

The book answer says 190 m/s.

Remember that energy is a scalar quantity, so we want to use the magnitude of the velocity as our $v_i$ and our $v_f$ instead of $(v_x)_i$, $(v_y)_i$ $(v_x)_f$ and $(v_y)_f$. Also, this equation may help you solve this problem: $\Delta{E} = E_f - E_i = 0J$ What are $E_f$ and $E_i$?

It also helps to solve your problem in terms of variables, check the units to see if they match up with velocity's units, and only then plug in given values! If you become competent at solving problems generally, it will trivialize solving problems individually. You may also recognize the equation you find when solving generally... :)