A projectile of mass 0.46 kg is shot from a cannon, at height 6.8 m, as shown in the figure, with an initial velocity vi having a horizontal component of 7.1m/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
Please tell us what is in the diagram.
black people
To find the solutions to the given questions, we can utilize the principles of projectile motion. Here's how we can find each answer:
1. To find the vertical component of the initial velocity at the end of the cannon's barrel, we need to determine the vertical velocity (Viy) at that point. Since the projectile is launched horizontally, the vertical component of the initial velocity is zero.
2. To find the maximum height the projectile achieves, we can use the equation:
∆y = Viy² / (2 * g)
Since the vertical component of the initial velocity is zero, the equation becomes:
∆y = 0 / (2 * g) = 0
This means that the maximum height (∆y) is 0 meters since the projectile does not rise above the cannon's barrel.
3. To find the magnitude of the velocity vector when the projectile hits the ground, we can break down the initial horizontal velocity (Vix) and the vertical acceleration (g) over the given horizontal distance (∆x) until the projectile hits the ground. We assume that air resistance is negligible.
We can use the equation:
∆x = Vix * t
In this case, ∆x is the given horizontal distance (∆x) past the end of the cannon's barrel. Since the initial horizontal velocity (Vix) is given (7.1 m/s), we can rearrange the equation to solve for time (t):
t = ∆x / Vix
Once we find the time, we can use it to determine the final vertical velocity (Vfy) using the equation:
Vfy = g * t
Finally, we can calculate the magnitude of the velocity vector using the Pythagorean theorem:
VF = √(Vix² + Vfy²)
4. To find the magnitude of the angle the projectile makes when impacting the ground, we can use the inverse tangent function (tan⁻¹) to find the angle (θ) using the final horizontal velocity (Vfx) and the final vertical velocity (Vfy) found in the previous step. The equation is:
θ = tan⁻¹(Vfy / Vfx)
5. To find the range of the projectile from the time it leaves the barrel until it hits the ground, we can use the equation:
Range = Vix * t
Here, t is found in the same way we calculated it in question 3.
By following these steps, you should be able to find the answers to each question using the given information.