Problem:
A cannonball was fired at an angle of 30°above the ground. If the muzzle’s velocity is 100m/s, determine the following:
a. How long it will stay in the air;
b. The maximum height it will reach;
c. The horizontal distance that it will travel; and
d. Its velocity just before it touches the ground
To solve this problem, we'll need to break it down into a few steps:
Step 1: Resolve the initial velocity into horizontal and vertical components.
Step 2: Use the vertical motion equation to find the time of flight.
Step 3: Use the time of flight to find the maximum height.
Step 4: Use the horizontal motion equation to find the horizontal distance traveled.
Step 5: Use the time of flight and the horizontal distance to find the final velocity.
Let's go through each step in detail:
Step 1: Resolving the initial velocity:
The initial velocity of the cannonball can be divided into horizontal and vertical components. The vertical component is given by V0y = V0 * sin(theta), where V0 is the initial velocity and theta is the angle of projection. The horizontal component is given by V0x = V0 * cos(theta).
In this case, V0 = 100 m/s and theta = 30°.
V0y = 100 * sin(30) = 50 m/s
V0x = 100 * cos(30) = 86.6 m/s
Step 2: Finding the time of flight:
The time of flight is the total time the cannonball stays in the air. We can use the vertical motion equation h = V0y * t - (1/2) * g * t^2, where h is the vertical displacement, V0y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the cannonball lands at the same height as it was launched, the vertical displacement h = 0. Rearranging the equation, we get -4.9 * t^2 + 50 * t = 0.
Solving this quadratic equation, we find t = 0 (which is the initial time) and t = 10.2041 s. The cannonball stays in the air for approximately 10.2 seconds.
Step 3: Finding the maximum height:
The maximum height is reached when the vertical velocity is zero. At the highest point, Vy = V0y - g * t = 0. Rearranging the equation, we get t = V0y / g.
Substituting the values, t = 50 / 9.8 = 5.1 s.
To find the maximum height H, we can use the vertical motion equation h = V0y * t - (1/2) * g * t^2.
H = 50 * 5.1 - 0.5 * 9.8 * (5.1)^2 = 255 - 127.755 = 127.245 m.
The maximum height reached by the cannonball is approximately 127.2 meters.
Step 4: Finding the horizontal distance traveled:
The horizontal distance traveled can be calculated using the horizontal motion equation d = V0x * t.
Substituting the values, d = 86.6 * 10.2 = 884.52 m.
The horizontal distance traveled by the cannonball is approximately 884.5 meters.
Step 5: Finding the final velocity:
The final velocity can be calculated using the horizontal motion equation v = V0x.
Substituting the value, v = 86.6 m/s.
The velocity just before the cannonball touches the ground is approximately 86.6 m/s.
a. The cannonball will stay in the air for approximately 10.2 seconds.
b. The maximum height reached by the cannonball is approximately 127.2 meters.
c. The horizontal distance traveled by the cannonball is approximately 884.5 meters.
d. The velocity just before the cannonball touches the ground is approximately 86.6 m/s.