whats the horizontal distance traveled by a model rocket launched with an intitial speed of 120 feet per second when the rocket is launched at an angle of 60 degrees
Projectile Motion
The ballistic analysis of a projectile may be simplified by combining the components of uniform and accelerated motion of the projectile. Consider a projectile fired from a gun with a muzzle velocity of V in a direction µº to the horizontal. We now ask ourselves, "How far from the gun will the projectile hit the ground?" and "How high will the projectile go?"
Breaking the velocity up into its horizontal and vertical components makes one thing immediately obvious. Only the vertical component is subject to the force of gravity. Applying the expression for accelerated motion, we can find the time of flight and the maximum height reached. The vertical component of the launch velocity is Vv = Vsin(µ). Therefore, from Vf = Vo - gt, the time to reach maximum height derives from 0 = Vsin(µ) - gt1 or Vsin(µ) = gt1 making
....... t1 = Vsin(µ)/g.
The height reached then derives from h = Vsin(µ)t1 - gt1^2/2 which can be written as h = gt1^2 - gt1^2/2 or
.......h = gt1^2/2.
The time for the projectile to return to the ground derives from
.......-h = -gt2^2/2. Since the height up equals the height down, it becomes clear that t1 = t2 making the total time of flight t = 2t1.
The horizontal distance traveled derives from d = Vcos(µ)t = 2Vt1cos(µ) = 2V[Vsin(µ)/g]cos(µ) making
.......d = V^2sin(2µ)/g.
Now, eliminating t1 from Vsin(µ) = gt1 and h = gt1^2/2, the maximum height reached becomes
.......h = V^2sin^2(µ)/2g.
As with uniform accelerated motion, these expressions ignore atmospheric drag.
To find the horizontal distance traveled by the model rocket, we can use the basic principles of projectile motion.
When a projectile is launched at an angle, we can break down its motion into horizontal and vertical components. In this case, the horizontal distance is what we are interested in.
The horizontal component of the initial velocity can be calculated by multiplying the initial velocity (120 feet per second) by the cosine of the launch angle (60 degrees).
So, the horizontal velocity (Vx) is given by:
Vx = Initial velocity * cos(angle)
Vx = 120 ft/s * cos(60°)
Vx = 120 ft/s * 0.5
Vx = 60 ft/s
Now, we need to determine the time of flight (t) for the rocket to reach the ground. The total time of flight can be determined using the vertical component of the initial velocity and the acceleration due to gravity.
The vertical component of the initial velocity can be calculated by multiplying the initial velocity (120 feet per second) by the sine of the launch angle (60 degrees).
So, the vertical velocity (Vy) is given by:
Vy = Initial velocity * sin(angle)
Vy = 120 ft/s * sin(60°)
Vy = 120 ft/s * √3/2
Vy = 60√3 ft/s
Now, we can use the formula for time of flight (t) for a projectile launched at an angle:
t = (2 * Vy) / g
Where g is the acceleration due to gravity (approximately 32.17 ft/s^2).
t = (2 * 60√3 ft/s) / 32.17 ft/s^2
t ≈ 3.49 seconds
Now that we have the time of flight, we can calculate the horizontal distance traveled (d) using the formula:
d = Vx * t
d = 60 ft/s * 3.49 s
d ≈ 209.4 feet
Therefore, when the rocket is launched at an angle of 60 degrees with an initial speed of 120 feet per second, the horizontal distance traveled by the rocket is approximately 209.4 feet.