A cannon is fired 30° above the horizontal with a velocity of 200 m/s from the edge of a 125 m high scliff calculate where the Cannonball lands on the plain below the answer is 3700
To solve this problem, we can use the kinematic equations of motion to determine the horizontal and vertical components of the cannonball's motion.
First, let's consider the vertical motion of the cannonball. The initial vertical velocity of the cannonball can be found by taking the sine of the launch angle:
Vyi = V * sinθ
Vyi = 200 * sin(30°)
Vyi = 100 m/s
Next, we can calculate the time it takes for the cannonball to reach the ground by using the following equation of motion:
h = Vyi * t + 0.5 * g * t^2
Where:
h = initial height of the cannonball = 125 m
g = acceleration due to gravity = 9.81 m/s^2
125 = 100 * t + 0.5 * 9.81 * t^2
125 = 100t + 4.905t^2
4.905t^2 + 100t - 125 = 0
Solving this quadratic equation will give us two possible solutions for t. We will discard the negative solution since time cannot be negative. The positive solution for t is approximately 5.01 seconds.
Once we have the time of flight, we can calculate the horizontal distance the cannonball travels by using the following equation:
dx = Vx * t
Where:
Vx = initial horizontal velocity of the cannonball = V * cosθ
Vx = 200 * cos(30°)
Vx = 173.21 m/s
dx = 173.21 * 5.01
dx = 868.24 m
Therefore, the cannonball lands on the plain below approximately 868.24 meters from the base of the cliff. However, based on the given answer of 3700, it seems there may have been an error in the calculations or information provided. Let me know if you need further assistance.