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
To solve these problems, we can use the principles of projectile motion. Projectile motion is the motion of an object in the air under the influence of gravity.
1. Find the vertical component of the initial velocity at the end of the cannon’s barrel:
The initial velocity (vi) has a horizontal component of 7.1 m/s. Since there is no vertical acceleration at the start, the vertical component of the initial velocity will remain unchanged. Therefore, the vertical component of the initial velocity is also 0 m/s.
2. Determine the maximum height the projectile achieves after leaving the end of the cannon's barrel:
We can use the kinematic equation for vertical displacement:
∆y = viy * t + (1/2) * a * t^2 ----(1)
where ∆y is the maximum height, viy is the vertical component of the initial velocity, t is the time taken to reach the maximum height, and a is the acceleration due to gravity (-9.8 m/s^2).
Since the projectile reaches its maximum height when its vertical velocity becomes zero, we can equate the vertical component of the initial velocity plus the product of acceleration due to gravity and time taken to reach the maximum height to zero:
viy + a * t = 0 ----(2)
From equation (2), we can solve for t:
t = -viy / a
Substituting this value of t in equation (1), we get:
∆y = -1/2 * (viy^2 / a)
Substituting viy = 0 and a = -9.8 m/s^2, we have:
∆y = -1/2 * (0^2 / -9.8)
∆y = 0
Therefore, the maximum height achieved by the projectile after leaving the end of the cannon's barrel is 0 meters.
3. Find the magnitude of the velocity vector when the projectile hits the ground:
The magnitude of the velocity vector is equal to the total velocity at any point, which is the vector sum of its horizontal and vertical components. To find the magnitude of velocity when the projectile hits the ground, we need to find the horizontal component of velocity (vix) and vertical component of velocity (viy) at that time.
The time taken for the projectile to reach the ground can be found using the equation for vertical displacement:
∆y = viy * t + (1/2) * a * t^2
Since the projectile starts and ends at the same height, ∆y is 0, and we can solve the equation for t.
0 = viy * t + (1/2) * (-9.8) * t^2
This equation is a quadratic equation in t. When solved, it provides two possible values for t (the time at which the projectile hits the ground). We will choose the positive value of t as the time represents a real-world situation.
Once we know t, we can calculate both vix and viy.
vix = vi * cos(θ)
viy = vi * sin(θ)
where θ is the angle with respect to the horizontal.
To find the magnitude of velocity when the projectile hits the ground, we can use the Pythagorean theorem:
|v| = sqrt(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 inverse tangent function:
θ = arctan(viy / vix)
5. Find the range of the projectile from the time it leaves the barrel until it hits the ground:
The range of a projectile is the horizontal distance it travels. In this case, the horizontal component of the initial velocity remains constant throughout the motion.
The time taken for the projectile to hit the ground can be found using the same method as described in question 3.
Once we know the time, we can calculate the horizontal distance traveled:
∆x = vix * t
where ∆x is the range and vix is the horizontal component of the initial velocity.
I hope this explanation helps you solve the given problems. Let me know if you have any further questions!