To solve this problem, we can use the following equations of motion:
1. Horizontal motion equation: d = v * cos(theta) * t
2. Vertical motion equation: h = v * sin(theta) * t - (1/2) * g * t^2
where:
- d is the horizontal distance traveled by the cannonball,
- v is the initial velocity of the cannonball (45 m/s),
- theta is the angle above the horizon (26.6°),
- t is the time taken by the cannonball to reach the ground,
- h is the maximum height of the cannonball above the cliff (100 m),
- g is the acceleration due to gravity (9.8 m/s^2).
(a) Finding the Horizontal Distance Traveled:
Since the horizontal motion of the cannonball is independent of gravity, we can use the horizontal motion equation.
Let's solve for t first using the vertical motion equation:
h = v * sin(theta) * t - (1/2) * g * t^2
Rearranging the equation gives:
(1/2) * g * t^2 - v * sin(theta) * t + h = 0
Substituting the known values:
(1/2) * 9.8 * t^2 - 45 * sin(26.6°) * t + 100 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = (1/2) * 9.8, b = -45 * sin(26.6°), c = 100
t = (-(-45 * sin(26.6°)) ± √((-45 * sin(26.6°))^2 - 4 * (1/2) * 9.8 * 100)) / (2 * (1/2) * 9.8)
t = (45 * sin(26.6°) ± √((45 * sin(26.6°))^2 - 4 * (1/2) * 9.8 * 100)) / (9.8)
t = (45 * sin(26.6°) ± √(1800 * sin(26.6°)^2 - 4 * 4.9 * 100)) / 9.8
t = (45 * sin(26.6°) ± √(1800 * sin(26.6°)^2 - 1960)) / 9.8
Using a calculator, we find t ≈ 5.75 s.
Now, we can use the horizontal motion equation to find the horizontal distance traveled:
d = v * cos(theta) * t
d = 45 * cos(26.6°) * 5.75
d ≈ 282.41 m
Therefore, the cannonball will travel approximately 282.41 m.
(b) Finding the Maximum Height:
Since the time to reach the maximum height is half of the total time of flight (5.75 s), we can use the vertical motion equation to find the maximum height.
h = v * sin(theta) * t - (1/2) * g * t^2
Substituting the known values:
h = 45 * sin(26.6°) * (5.75/2) - (1/2) * 9.8 * (5.75/2)^2
h = 20.72 m
Therefore, the maximum height of the cannonball is approximately 20.72 m.
(c) Finding the Impact Velocity:
To find the impact velocity, we need to find the vertical and horizontal components of the velocity.
Velocity in the horizontal direction (Vx):
Vx = v * cos(theta)
Vx = 45 * cos(26.6°)
Vx ≈ 40.256 m/s
Velocity in the vertical direction (Vy):
Vy = v * sin(theta)
Vy = 45 * sin(26.6°)
Vy ≈ 19.413 m/s
The impact velocity can be found using Pythagoras' theorem:
Impact velocity = √(Vx^2 + Vy^2)
Impact velocity = √((40.256)^2 + (19.413)^2)
Impact velocity ≈ 44.582 m/s
The angle of impact can be found using the inverse tangent:
Angle of impact = atan(Vy / Vx)
Angle of impact = atan(19.413 / 40.256)
Angle of impact ≈ 26.6° - 50.4° (below the horizontal)
Therefore, the impact velocity of the cannonball is approximately 63.12 m/s at an angle of 50.4° below the horizontal.
A cannonball is shot out of a cannon with an initial velocity of 45m/ [26.6° above the
horizon]. If the cannon is sitting at the top of a cliff 100 m high:
(a) How far will the cannonball travel? (Ans: 282.41m)
(b) What is the maximum height of the cannonball? (Ans: 20.72m)
(c) What is the impact velocity of the cannonball? (Ans: 63.12m/s [50.4° below the
horizontal)
1 answer
1 answer