PLEASE HELP IM LOST
A projectile of mass 0.653 kg is shot from a cannon. The end of the cannon's barrel is at height 6.5 m, as shown in the figure. The initial velocity of the projectile is 8.5 m/s. The projectile rises to a maximum height of Delta Y above the end of the cannon's barrel and strikes the ground a horizontal distance Delta X past the end of the cannon's barrel.
Determine the vertical component of the
initial velocity at the end of the cannon's bar rel, where the projectile began its trajectory. The acceleration of gravity is 9.8 m/s. Answer in units of m/s.
Part 2
Determine the maximum height Delta y the projectile achieves after leaving the end of the cannon's barrel. Answer in units of m.
Part 3
Determine the time it takes for the projectile to reach its maximum height. Answer in units of s.
Part 4
How long does it take the projectile to hit the ground?
Answer in units of s.
To solve this problem, we can use the equations of motion for projectile motion.
Part 1:
The vertical component of the initial velocity can be found using the equation v_y = v_initial * sin(theta), where v_initial is the initial velocity and theta is the angle at which the projectile is launched. In this case, the projectile is launched horizontally, so the angle theta is 0 degrees. Thus, the vertical component of the initial velocity is:
v_y = 8.5 m/s * sin(0) = 0 m/s
Therefore, the vertical component of the initial velocity at the end of the cannon's barrel is 0 m/s.
Part 2:
The maximum height reached by the projectile can be found using the equation Delta y = (v_initial^2 * sin^2(theta)) / (2 * g), where Delta y is the maximum height, v_initial is the initial velocity, theta is the angle at which the projectile is launched, and g is the acceleration due to gravity. In this case, the angle theta is 0 degrees, so sin^2(theta) = sin^2(0) = 0. Therefore, the maximum height is:
Delta y = (8.5 m/s)^2 * sin^2(0) / (2 * 9.8 m/s^2) = 0 m
Thus, the maximum height achieved by the projectile is 0 m.
Part 3:
The time it takes for the projectile to reach its maximum height can be found using the equation t = v_initial * sin(theta) / g. In this case, the angle theta is 0 degrees and the vertical component of the initial velocity is 0 m/s. Therefore, the time is:
t = 0 m/s / 9.8 m/s^2 = 0 s
Thus, it takes 0 seconds for the projectile to reach its maximum height.
Part 4:
The time it takes for the projectile to hit the ground can be found using the equation t = 2 * v_initial * sin(theta) / g. In this case, the angle theta is 0 degrees. Therefore, the time is:
t = 2 * 8.5 m/s * sin(0) / 9.8 m/s^2 = 0 s
Thus, it takes 0 seconds for the projectile to hit the ground.
To solve this problem, let's break it down into parts and use the principles of projectile motion.
1. Determine the vertical component of the initial velocity at the end of the cannon's barrel:
The vertical component of the initial velocity can be found using the formula: Vy = V * sin(θ), where V is the initial velocity and θ is the angle of launch. In this case, the angle is not given, so we assume it to be 90 degrees because the projectile is being launched vertically. Therefore, sin(90) = 1. Plug in the values to calculate Vy:
Vy = 8.5 m/s * sin(90) = 8.5 m/s
2. Determine the maximum height Delta y the projectile achieves after leaving the end of the cannon's barrel:
To find the maximum height, we can use kinematic equations. The equation we'll use is: Δy = Vy^2 / (2g), where Δy is the maximum height, Vy is the vertical component of initial velocity, and g is the acceleration due to gravity (9.8 m/s^2).
Plug in the values to calculate Δy:
Δy = (8.5 m/s)^2 / (2 * 9.8 m/s^2) = 3.650 m
3. Determine the time it takes for the projectile to reach its maximum height:
To find the time, we can again use kinematic equations. We'll use the following equation: Vy = V0y - gt, where V0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time taken.
Rearrange the equation to solve for t:
t = (V0y - Vy) / g
Plug in the values to calculate t:
t = (8.5 m/s - (-8.5 m/s)) / 9.8 m/s^2 = 1.73 s
4. Determine how long it takes the projectile to hit the ground:
To determine this, we'll use the equation: Δy = V0y * t - 0.5 * g * t^2, where Δy is the maximum height (which we found earlier), V0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time taken.
Rearrange the equation to solve for t:
0 = V0y * t - 0.5 * g * t^2
This is a quadratic equation, and we solve it using the quadratic formula. However, in this case, we can simply take the positive root since we're interested in the time it takes for the projectile to hit the ground.
Plug in the values to calculate t:
0 = 8.5 m/s * t - 0.5 * 9.8 m/s^2 * t^2
Simplifying the equation, we get:
4.9 t^2 - 8.5 t = 0
t(4.9 t - 8.5) = 0
t = 0 (which we discard) or t = 8.5 m/s / 4.9 m/s^2 = 1.7351 s (rounded to 4 decimal places)
Therefore, the answers are:
- The vertical component of the initial velocity at the end of the cannon's barrel is 8.5 m/s.
- The maximum height Delta y the projectile achieves after leaving the end of the cannon's barrel is 3.650 m.
- The time it takes for the projectile to reach its maximum height is 1.73 s.
- The time it takes the projectile to hit the ground is 1.7351 s.