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
It seems to me that the angle needs to be known. Is that in the diagram?
To solve these questions, we need to apply the equations of motion for projectile motion. Here's how to solve each part:
Let's break down the given information:
Mass of projectile (m) = 0.46 kg
Height of cannon (h) = 6.8 m
Horizontal component of initial velocity (vi) = 7.1 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
1. Find the vertical component of the initial velocity (Viy):
The initial velocity (vi) can be divided into horizontal and vertical components using trigonometry. Since the horizontal component is given (7.1 m/s), we need to find the vertical component.
Viy = vi * sin(theta), where theta is the angle of the initial velocity with respect to the horizontal.
To find theta, we can use the fact that Viy/Vix = tan(theta), where Vix is the horizontal component of velocity (given).
theta = arctan(Viy/Vix)
Given that Vix = 7.1 m/s and Viy/Vix = tan(theta), we can solve for theta.
Once we have theta, we can calculate Viy using Viy = vi * sin(theta).
2. Determine the maximum height achieved by the projectile (Δy):
Since the projectile rises to a maximum height, its vertical velocity at the maximum height (Vy_max) is zero.
Using the equation Vy_max = Viy_max - g * t_max, where t_max is the time taken to reach the maximum height, we can solve for t_max.
Now, we can calculate Δy using the equation Δy = Viy * t_max - (1/2) * g * t_max^2.
3. Find the magnitude of the velocity vector when the projectile hits the ground:
The velocity vector when the projectile hits the ground is the resultant of the horizontal and vertical components.
To find the magnitude, we use the equation v^2 = Vix^2 + Viy^2.
4. Find the magnitude of the angle (with respect to horizontal) the projectile makes when impacting the ground:
The angle with respect to the horizontal can be found using the equation theta = arctan(Viy/Vix).
5. Find the range of the projectile from the time it leaves the barrel until it hits the ground:
The range is the horizontal distance traveled by the projectile.
Using the equation Δx = Vix * t, where t is the time of flight, we can find the range.
Now we have the steps to solve each part of the problem. You can use the given values and apply these equations to find the answers.