Brandon launches a projectile at an angle of 75 above the horizontal, which strikes he ground a certain distance down range. fo what oher angle of launch at the same speed would the projectile land just as far away?
d = V^2(sin(2µ))/g
Since the maximum range occurs at a launch angle of 45º, launch angles of +/- equal increments of µ on either side of 45º will result in the same horizontal distance traveled.
Therefore, the launch angle required to result in the same range of the projectile
becomes 45 - (75 - 30) = 15º.
Vsin(150) = Vsin30 = .5V
To find the angle of launch that would make the projectile land just as far away, we need to understand the physics behind projectile motion.
The motion of a projectile can be analyzed by breaking it into horizontal and vertical components. The horizontal component determines the distance traveled, while the vertical component affects the time of flight and the height reached.
Let's denote the original angle of launch as θ_1 and the distance traveled downrange as d.
Given:
Angle of launch (θ_1) = 75°
To find the angle of launch (θ_2) that will give the same downrange distance, we can use the principle of symmetry in projectile motion. The maximum range occurs when the projectile is launched at an angle of 45°. Since the d value is the same for both scenarios, we can find the angle θ_2 that results in the same range.
To calculate this new angle, we can use the following steps:
1. Determine the vertical component of velocity (v_y) at the original angle of launch (θ_1).
- Since the projectile is launched at an angle, we need to find the vertical component of its initial velocity.
- v_y = v_initial * sin(θ_1)
2. Use the principle of symmetry to find the angle θ_2 that will give the same downrange distance.
- The angle of 45° gives the maximum range, so the total angle of ascent and descent will be 90°.
- Therefore, the angle of descent is (90° - θ_2).
- Since the vertical component of velocity at the apex is zero, v_y at the point of descent is v_initial * sin(θ_2).
3. Solve for θ_2 using the fact that the vertical component of velocity at the point of descent is the same as at launch.
- v_y at the point of descent = v_y at launch
- v_initial * sin(θ_2) = v_initial * sin(θ_1)
4. Solve for θ_2 by canceling the common terms and isolating the angle.
- sin(θ_2) = sin(θ_1)
- θ_2 = sin^(-1)(sin(θ_1))
Using the given angle of launch (θ_1 = 75°), we can calculate the angle θ_2.
θ_2 = sin^(-1)(sin(75°))
Evaluating this expression, we have:
θ_2 ≈ 75°
So, launching the projectile at an angle of approximately 75° would make it land just as far away.