derive the expression for time of flight maximum height & range when a body is projected at an angle thita frommthe vertical position with a proper diagram.
To derive the expressions for time of flight, maximum height, and range of a projectile launched at an angle θ from the vertical position, we can use the equations of motion and principles of projectile motion. Let me guide you step by step:
1. Consider the initial conditions and coordinate system:
- Assume the projectile is launched from the ground level.
- Take the upward direction as positive.
- The initial velocity can be broken down into horizontal and vertical components, given as:
- Initial vertical velocity (Vy₀) = v₀ * sin(θ)
- Initial horizontal velocity (Vx₀) = v₀ * cos(θ)
(Here, v₀ represents the initial magnitude of the velocity.)
Now, let's begin with the different parameters:
2. Time of Flight (T):
- The time of flight is the total duration for which the projectile remains in the air.
- Considering the vertical motion, at the maximum height, the projectile reaches its highest point where the vertical component of velocity (Vy) becomes zero.
- The time taken to reach this point can be found using the equation:
- Vy = Vy₀ + a * t
- Since Vy = 0 at the highest point, we can solve for the time taken (t₁) to reach the maximum height:
0 = Vy₀ + (-g) * t₁ [as acceleration in the vertical direction is -g, the acceleration due to gravity]
t₁ = Vy₀ / g
- Next, to find the total time of flight (T), we consider the time taken for the projectile to reach the highest point and then return to the ground. Since the time taken to reach the highest point is the same as the time taken to return to the ground, we have:
- T = 2 * t₁
- T = 2 * (Vy₀ / g)
3. Maximum Height (H):
- The maximum height reached by the projectile can be determined using the equation of motion:
- Vy = Vy₀ + a * t
- Vy = 0 at the highest point, so:
0 = Vy₀ + (-g) * t₂ [t₂ is the time taken to reach the highest point]
t₂ = Vy₀ / g
- Now, using the equation of motion in the vertical direction:
- H = Sy - Sy₀
- Here, Sy = Vy₀ * t₂ + (1/2) * (-g) * t₂² = (Vy₀²) / (2 * g)
[Using the equation Sy = Sy₀ + Vy₀ * t + (1/2) * a * t², where Sy₀ = 0 (initial height) and acceleration (a) is -g.]
4. Range (R):
- The range is the horizontal distance covered by the projectile.
- The horizontal distance can be found using the equation:
- R = Vx * T
- Since Vx = Vx₀ throughout the motion, we get:
R = Vx₀ * T
R = v₀ * cos(θ) * (2 * (Vy₀ / g))
With these equations, you have derived the expressions for the time of flight (T), maximum height (H), and range (R) of the projectile launched at an angle θ from the vertical position.