A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal range d.
(a) At what angle θ is the rock thrown?
75
To determine the angle θ at which the rock is thrown, we can use the concept of projectile motion. In this case, we know that the maximum height of the rock's flight is equal to its horizontal range.
Here's how you can approach solving this problem:
1. Write down the given information:
- Maximum height = Horizontal range (h = d)
2. Recall the equations of motion for projectile motion:
- Vertical displacement (h) = u^2 * sin^2(θ) / (2 * g)
- Horizontal displacement (d) = u^2 * sin(2θ) / g
3. Since the maximum height is equal to the horizontal range, we can equate the two equations:
=> u^2 * sin^2(θ) / (2 * g) = u^2 * sin(2θ) / g
4. Cancel out the common terms and simplify the equation:
=> sin^2(θ) / (2 * g) = sin(2θ) / g
5. Manipulate the equation to isolate the angle θ:
=> sin^2(θ) = sin(2θ) / 2
6. Use a trigonometric identity to simplify the equation further:
=> sin^2(θ) = 2 * sin(θ) * cos(θ)
7. Divide both sides of the equation by sin(θ):
=> sin(θ) = 2 * cos(θ)
8. Divide both sides of the equation by cos(θ):
=> tan(θ) = 2
9. Take the inverse tangent (arctan) of both sides to find the value of θ:
=> θ = arctan(2)
Therefore, the angle at which the rock is thrown (θ) is equal to arctan(2), approximately 63.4 degrees.