A cannon fires a free projectile at an angle 85 from the ground. The cannonball has a speed of 150 m/s and it is launched from a hill 20 m above the plain where it will land
1. What is the x component of the cannonballs initial velocity
2. What is the y component of the cannonballs initial velocity
3. What is the x component of the cannonballs velocity at the peak of its trajectory?
4. What is the y component of the cannonballs velocity at the peak of its trajectory?
5. What is the cannonballs acceleration at the peak of its trajectory?
6. What is the height above the plain of the cannonball at the peak of its trajectory?
7. How long will the cannonball take to reach the peak of its trajectory?
8. How long will the cannonball take to fall from the ground from the peak?
9. What is the total time the cannonball will spend in the air?
10. What will be the y component of the final velocity of the cannonball when it hits the ground?
11. What is the total magnitude of the final velocity of the cannonball when it hits the ground?
12. What is the range of the cannonball?
13. At what angle will the cannonball hit the ground?
Vo = 150m/s[85o]
1. Xo = 150*cos85 = 13.07 m/s.
2. Yo = 150*sin85 = 149.4 m/s.
3. X = Xo
4. Y = 0
5. a = g = 9.8 m/s^2.
6. h = ho + (Y^2-Yo^2)/2g
h = 20 + (0-149.4^2)/-19.6 = 1159 m.
Above the plain.
7. Y = Yo + g*t = 0 @ peak.
149.4 -9.8t = 0
9.8t = 149.4
Tr = 15.2 s. = Rise time.
8. h = 0.5g*t^2 = 1159 m.
4.9*t^2 = 1159
t^2 = 236.5
Tf = 15.38 s. = Fall time.
9. T = Tr+Tf = 15.2 + 15.38 = 30.58 s.
10. Y=Yo + g*t = 0 + 9.8*15.38=150.7 m/s
11. V = sqrt(Xo^2+Y^2)
V = sqrt(13.07^2 + 150.7^2) = 151.3 m/s.
12. Range = Xo*(Tr+Tf)
Range = 13.07 * (15.2 + 15.38) = 400 m.
13. Tan A=Y/Xo = 150.7/13.07 = 11.53022
A = 85.0o
To solve these questions, we will use the equations of motion for projectile motion. In projectile motion, an object moves in a curved path under the influence of gravity.
1. The x-component of the initial velocity can be found using the equation:
Vx = V * cos(θ)
where V is the magnitude of the initial velocity and θ is the launch angle.
Substituting the given values:
Vx = 150 m/s * cos(85°)
Vx ≈ -25.52 m/s
2. The y-component of the initial velocity can be found using the equation:
Vy = V * sin(θ)
Substituting the given values:
Vy = 150 m/s * sin(85°)
Vy ≈ 147.68 m/s
3. The x-component of the velocity at the peak of the trajectory is zero since there is no horizontal acceleration.
4. The y-component of the velocity at the peak of the trajectory can be found by using the equation:
Vy_peak = Vy_initial - g * t
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time taken to reach the peak (which we will calculate in question 7).
Substituting the values:
Vy_peak = 147.68 m/s - 9.8 m/s^2 * t (where t is the time to reach the peak)
5. The acceleration of the cannonball at the peak of its trajectory is -g, which is equal to -9.8 m/s^2. The acceleration is negative because it acts in the opposite direction of motion.
6. The height above the plain of the cannonball at the peak of its trajectory is given by the y-component of the initial position, which is 20 m.
7. To find the time taken to reach the peak of the trajectory, we can use the equation:
Vy_peak = Vy_initial - g * t
Rearranging and substituting the values:
0 = 147.68 m/s - 9.8 m/s^2 * t
Solving for t:
t ≈ 15.07 s (rounded to two decimal places)
8. The time taken to fall from the peak to the ground is the same as the time taken to reach the peak, so it is approximately 15.07 seconds.
9. The total time the cannonball spends in the air is given by twice the time taken to reach the peak:
Total time = 2 * t
Substituting the value of t:
Total time ≈ 2 * 15.07 s
Total time ≈ 30.14 s
10. The y-component of the final velocity of the cannonball when it hits the ground can be found using the same equation as in question 4:
Vy_final = Vy_initial - g * t
Substituting the values:
Vy_final = 147.68 m/s - 9.8 m/s^2 * 30.14 s
Vy_final ≈ -295.24 m/s
11. The total magnitude of the final velocity of the cannonball when it hits the ground can be found using the Pythagorean theorem:
V_final = sqrt(Vx_final^2 + Vy_final^2)
Substituting the values:
V_final = sqrt((-25.52 m/s)^2 + (-295.24 m/s)^2)
V_final ≈ 296.9 m/s (rounded to one decimal place)
12. The range of the cannonball is given by the equation:
R = Vx_initial * time of flight
Substituting the values:
R = -25.52 m/s * 30.14 s
R ≈ -769.5 m (rounded to one decimal place)
Since range is a scalar quantity, the negative sign indicates the opposite direction of the initial velocity.
13. To find the angle at which the cannonball hits the ground, we can use the equation:
tan(θ) = (Vy_final) / (Vx_final)
Substituting the values:
tan(θ) = (-295.24 m/s) / (-25.52 m/s)
θ ≈ 84.9° (rounded to one decimal place)
By following these steps, you should be able to answer each of the given questions relating to the cannonball fired at an angle from a hill.