A rocket is fired with an initial velocity of 200m/s at an angle of 60' to the horizontal.Find (1)the maximum height reached.
To find the maximum height reached by the rocket, we need to use the equations of motion.
The vertical motion of the rocket can be described by the following equations:
1. The vertical displacement formula:
Δy = (v₀y * t) + (1/2 * a * t²)
2. The vertical velocity formula:
v_y = v₀y + (a * t)
Where:
Δy = vertical displacement or maximum height
v₀y = initial vertical velocity
t = time
a = acceleration due to gravity (approximately -9.8 m/s²)
Given:
Initial velocity (v₀) = 200 m/s
Launch angle = 60 degrees
First, we need to find the initial vertical velocity (v₀y). Since the rocket is launched at an angle of 60 degrees to the horizontal, we need to decompose the initial velocity into its horizontal (v₀x) and vertical (v₀y) components.
v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)
where θ is the launch angle.
Let's calculate v₀y:
v₀y = 200 m/s * sin(60°)
= 200 m/s * √3/2
= 100√3 m/s
Now, let's find the time (t) it takes for the rocket to reach its maximum height. At the maximum height, the vertical velocity will be zero since the rocket momentarily stops before falling back down.
Using the vertical velocity formula:
v_y = v₀y + (a * t)
0 = 100√3 m/s + (-9.8 m/s² * t)
Solving for t:
9.8 m/s² * t = 100√3 m/s
t = (100√3 m/s) / 9.8 m/s²
t ≈ 3.20 s
Now that we have the time (t), we can calculate the maximum height (Δy) using the vertical displacement formula:
Δy = (v₀y * t) + (1/2 * a * t²)
Δy = (100√3 m/s * 3.20 s) + (1/2 * -9.8 m/s² * (3.20 s)²)
= (320√3 m²/s) - (1/2 * 9.8 m/s² * 10.24 s²)
≈ 184.4 m
Therefore, the maximum height reached by the rocket is approximately 184.4 meters.
To find the maximum height reached by the rocket, we can use the equations of projectile motion. Here's how to approach the problem step by step:
Step 1: Split the initial velocity into its horizontal and vertical components.
Given the initial velocity (v) of the rocket is 200 m/s at an angle of 60 degrees to the horizontal, we can split it into its horizontal and vertical components.
The horizontal component (v_x) can be found using the equation:
v_x = v * cos(theta)
where v is the initial velocity and theta is the angle of launch.
Substituting the values, we get:
v_x = 200 * cos(60)
v_x ≈ 200 * 0.5
v_x = 100 m/s
The vertical component (v_y) can be found using the equation:
v_y = v * sin(theta)
Substituting the values, we get:
v_y = 200 * sin(60)
v_y ≈ 200 * 0.866
v_y ≈ 173.2 m/s
Step 2: Calculate the time it takes to reach the maximum height.
At the maximum height, the vertical component of velocity becomes zero. We can calculate the time (t) it takes for v_y to change from 173.2 m/s to zero using the following equation:
0 = v_y + g * t
where g is the acceleration due to gravity (~9.8 m/s^2).
Rearranging the equation, we get:
t = -v_y / g
Substituting the values, we get:
t = -173.2 / 9.8
t ≈ -17.66 s (Note: The negative sign indicates opposite direction to the chosen reference frame. We take the magnitude here.)
Step 3: Calculate the maximum height (H).
To find the maximum height, we can use the equation:
H = v_y^2 / (2 * g)
Substituting the values, we get:
H = (173.2^2) / (2 * 9.8)
H ≈ 150743.28 / 19.6
H ≈ 7679.52 m
Therefore, the maximum height reached by the rocket is approximately 7679.52 meters.
Note: It's important to note that this solution assumes no air resistance and a constant acceleration due to gravity.
initial Kinetic energy in the upward velocity component: 1/2 m (200sin60)^2
final PE at top: mg height
set them equal, solve for height.