A projectile of mass 0.377 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 6.1 m/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. The cannon is 49 degrees above the ground.
Determine the maximum height the projectile
achieves after leaving the end of the cannon’s
barrel.
Find the magnitude of the velocity vector
when the projectile hits the ground.
Find the magnitude of the angle (with respect
to horizontal) the projectile makes when im-
pacting the ground.
Find the range of the projectile from the time
it leaves the barrel until it hits the ground.
To solve these problems, we can use the principles of projectile motion. Here's how to approach each question:
1. Determine the maximum height the projectile achieves after leaving the end of the cannon’s barrel:
To find the maximum height, we need to find the vertical component of the initial velocity. We know the initial velocity's horizontal component is 6.1 m/s, and the launch angle is 49 degrees above the ground. To find the vertical component, we can use the equation:
v_y = vi * sin(theta)
where v_y is the vertical component of the velocity and theta is the launch angle. Calculate the value of v_y using the given values, and then we can use the vertical motion equation:
y = yi + v_y * t - 0.5 * g * t^2
where y is the maximum height, yi is the initial vertical position (6.6 m), t is the time, and g is the acceleration due to gravity (9.8 m/s^2).
2. Find the magnitude of the velocity vector when the projectile hits the ground:
The magnitude of the velocity vector is the total velocity of the projectile. To find it, we can use the horizontal and vertical components of the velocity. The horizontal component is given as 6.1 m/s, and the vertical component can be found using the formula mentioned earlier. We can then calculate the magnitude of the velocity vector using the Pythagorean theorem:
|v| = sqrt(v_x^2 + v_y^2)
where v is the magnitude of the velocity vector, v_x is the horizontal component of the velocity, and v_y is the vertical component of the velocity.
3. Find the magnitude of the angle (with respect to horizontal) the projectile makes when impacting the ground:
To find the angle, we need the horizontal and vertical components of the velocity when the projectile hits the ground. The horizontal component is given as 6.1 m/s, and the vertical component can be found using the formula mentioned earlier. We can use the inverse tangent function to find the angle:
angle = arctan(v_y / v_x)
where angle is the magnitude of the angle with respect to the horizontal.
4. 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. To find it, we need the horizontal component of the velocity and the time of flight. The horizontal component is given as 6.1 m/s, and the time of flight can be calculated using the equation:
t = 2 * v_y / g
where t is the time of flight. Finally, we can calculate the range:
range = v_x * t
By using these formulas and the given values, you can solve each question and find the answers.