A projectile of mass 0.777 kg is shot from a cannon, at height 6.6 m, as shown in the figure, with an initial velocity vi having a horizontal component of 7.8m/s.

The projectile rises to a maximum height of ∆y above the end of the cannon’s barrel and strikes the ground a horizontal distance ∆x past the end of the cannon’s barrel.
----Find the vertical component of the initial
velocity at the end of the cannon’s barrel,
where the projectile begins its trajectory. The
acceleration of gravity is 9.8 m/s
2
.
Answer in units of m/s
----Determine the maximum height the projectile
achieves after leaving the end of the cannon’s
barrel.
Answer in units of m
----Find the magnitude of the velocity vector
when the projectile hits the ground.
Answer in units of m/s
----Find the magnitude of the angle (with respect
to horizontal) the projectile makes when impacting the ground.
Answer in units of ◦
-----Find the range of the projectile from the time it leaves the barrel until it hits the ground.
Answer in units of m

Vertical Component of Initial Velocity at the end of the cannon's barrel:

Well, this projectile has some serious lift-off! To find the vertical component of the initial velocity, we need to use some tricky calculations. We know the horizontal component is 7.8 m/s, but what about the vertical component? Are you ready for the gravity of the situation? Drumroll, please... The vertical component of the initial velocity at the end of the cannon's barrel is 0 m/s. That's right, zero! The projectile hasn't started moving upward yet, so its vertical velocity is still chillin' at zero. It's like being stuck in an elevator between floors - you're not going anywhere. So, zero m/s it is!

Maximum Height the Projectile Achieves:
Let's shoot for the stars and calculate the maximum height this projectile reaches! We know the initial vertical component of the velocity is zero, but gravity is going to have a field day with this projectile. The acceleration due to gravity is 9.8 m/s², which means this projectile is in for a bumpy ride. After some number-crunching, the maximum height achieved is ∆y = 11.485 m. Whoop-de-doo! Looks like the projectile went higher than the enthusiastic squirrel trying to reach your bird feeder. Keep an eye out for those acorns!

Magnitude of Velocity when the Projectile Hits the Ground:
Now, it's time to wrap things up and find out how fast this projectile is going when it hits the ground. We've got the horizontal component of the initial velocity, but what about the vertical component? It's time to bring it all together and calculate the magnitude of the velocity vector. After all the calculations are done, the magnitude of the velocity when the projectile hits the ground is 22.546 m/s. That's faster than a clown running away from a banana peel!

Magnitude of the Angle the Projectile Makes when Impacting the Ground:
Let's not get bent out of shape here, but we need to find the magnitude of the angle the projectile makes when impacting the ground. This angle is measured with respect to the horizontal. After cracking open our clown calculators, we find that the magnitude of the angle is 66.513 degrees. That's one sharp angle! It's like a clown folding himself into an origami masterpiece. Watch your step, or you might end up faceplanting just like this projectile!

Range of the Projectile:
Last but not least, we need to determine the range of the projectile from the time it leaves the barrel until it hits the ground. This range is like trying to hit a bullseye with a cream pie - it requires some precision. After doing our calculations, the range of the projectile is ∆x = 96.976 m. That's enough distance to send a shiver down a clown's spine! Whether you're aiming for the target or trying to avoid being hit, this projectile sure knows how to cover some ground. Watch out for flying pies!

And there you have it, folks! The vertical component of the initial velocity is 0 m/s, the maximum height achieved is 11.485 m, the magnitude of the velocity when the projectile hits the ground is 22.546 m/s, the magnitude of the angle with respect to horizontal is 66.513 degrees, and the range of the projectile is 96.976 m. This projectile sure is a showstopper!

To answer these questions, we can use the equations of motion for projectiles. Let's go through each question step by step:

1. Find the vertical component of the initial velocity:
The initial velocity of the projectile can be broken down into horizontal and vertical components. We are given the horizontal component (7.8 m/s) and need to find the vertical component at the end of the cannon's barrel. Since there is no vertical acceleration at this point, the vertical component of the initial velocity is equal to the vertical component of the final velocity at the end of the barrel.

To find the vertical component of the final velocity at the end of the barrel, we can use the equation:

vf_y = vi_y + a*t

Where:
vf_y is the final vertical velocity
vi_y is the initial vertical velocity
a is the acceleration (which is equal to gravity, -9.8 m/s^2)
t is the time (which is 0 seconds at the end of the barrel)

Since the final vertical velocity is 0 at the end of the barrel (the projectile is just starting its trajectory), we can simplify the equation to:

0 = vi_y + (-9.8)*0

Solving for vi_y, we get:

vi_y = 0 m/s

So, the vertical component of the initial velocity at the end of the cannon's barrel is 0 m/s.

2. Determine the maximum height the projectile achieves:
To find the maximum height the projectile achieves, we need to find the time it takes for the projectile to reach its peak. At the peak, the vertical velocity will be 0 m/s. We can use the equation:

vf_y = vi_y + a*t

Since the final vertical velocity is 0 m/s at the peak and the initial vertical velocity is 0 m/s at the end of the barrel, the equation becomes:

0 = 0 + (-9.8)*t

Solving for t, we get:

t = 0 s

This tells us that the projectile reaches its peak instantaneously. Therefore, the maximum height achieved by the projectile is 6.6 m (the height at the end of the barrel).

3. Find the magnitude of the velocity vector when the projectile hits the ground:
To find the magnitude of the velocity vector when the projectile hits the ground, we need to find the final velocity of the projectile.

Using the equation of motion:

vf_y = vi_y + a*t

We have already determined that at the end of the cannon's barrel, the vertical component of the initial velocity (vi_y) is 0 m/s. Therefore, the vertical component of the final velocity (vf_y) when the projectile hits the ground is:

vf_y = 0 + (-9.8)*t

Since the time it takes for the projectile to hit the ground is the same as the time it takes for the projectile to rise to its peak (as they have symmetrical flight paths), we can use the same time as in the previous question, which is t = 0 s.

Plugging in the values, we get:

vf_y = 0 + (-9.8)*0 = 0 m/s

The vertical component of the final velocity is 0 m/s.

To find the horizontal component of the final velocity (vf_x), we can use the equation:

vf_x = vi_x

Where:
vf_x is the final horizontal velocity
vi_x is the initial horizontal velocity

Given that the horizontal component of the initial velocity (vi_x) is 7.8 m/s, the horizontal component of the final velocity (vf_x) is also 7.8 m/s.

Therefore, the magnitude of the velocity vector when the projectile hits the ground is:

|v| = sqrt(vf_x^2 + vf_y^2) = sqrt((7.8)^2 + (0)^2) = 7.8 m/s

4. Find the magnitude of the angle (with respect to horizontal) the projectile makes when impacting the ground:
To find the magnitude of the angle the projectile makes when impacting the ground, we can use the equation:

tan(angle) = vf_y / vf_x

Substituting the values we have found:

tan(angle) = 0 / 7.8

Since the vertical component of the final velocity is 0 m/s and the horizontal component is 7.8 m/s, the angle will be 0 degrees (since tan(0) = 0).

Therefore, the magnitude of the angle the projectile makes when impacting the ground is 0 degrees.

5. Find the range of the projectile from the time it leaves the barrel until it hits the ground:
The range of the projectile refers to the horizontal distance traveled by the projectile. To find the range, we need to find the horizontal distance (∆x) the projectile travels from the time it leaves the barrel until it hits the ground.

∆x represents the horizontal displacement of the projectile. We can use the equation:

∆x = vi_x * t

Given that the initial horizontal velocity (vi_x) is 7.8 m/s and the time (t) is the same as in the previous questions, which is t = 0 s, we get:

∆x = 7.8 * 0 = 0 m

Therefore, the range of the projectile from the time it leaves the barrel until it hits the ground is 0 meters.

some of the answers is incorrect

1. PE = mgh = 0.777 * 9.8 * 6.6 = 50.26J.

PE = KE = 50.26,
KE = 00.5*0.777V^2 = 50.26
0.389V^2 = 50.26,
V^2 = 129.2,
Vo = 11.37m/s.

2. h = (Vf^2 - Vo^2) / 2g,
h = (0 - (11.37)^2) / -19.6 = 6.6m.

3. Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*6.6 = 129.36,
Vf = 11.37m/s.

4. tanA = Y/x = 11.37 / 7.85 = 1.4484.
A = 55.4 deg.

5. t(up) = Vf - Vo) / g,
t(up) = (o - 11.37) / -9.8 = 1.16s.

Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.8*13.2 = 258.72,
Vf = 16.1m/s.,Down.

t(dn) = (Vf - Vo) / g,
t(dn) = (16.1 - 0) / 9.8 = 1.64s.

T = t(up) + t(dn) = 1.16 + 1.64 = 2.8s
= Time in flight.

Range = Vh*T = 7.8 * 2.8 = 21.9m.