A projectile’s launch speed is five times its speed at maxi-mum height. Find launch angle θ
Question
A projectile’s launch speed is five times its speed at maxi-mum height. Find launch angle θ
Solution
The speed of a projectile at its maximum height is solely its horizontal component because the vertical component becomes zero at the maximum height.
Let's denote:
- V0 as the launch speed
- Vh as the horizontal speed
- Vv as the vertical speed
- θ as the launch angle
From the problem, we know that V0 = 5 * Vh.
We also know from physics that:
- Vh = V0 * cos(θ)
- Vv = V0 * sin(θ)
At the maximum height, Vv = 0, so the speed of the projectile is equal to Vh.
Substituting Vh = V0 * cos(θ) into the equation V0 = 5 * Vh, we get:
V0 = 5 * V0 * cos(θ)
Dividing both sides by V0 (assuming V0 ≠ 0), we get:
1 = 5 * cos(θ)
Solving for cos(θ), we get:
cos(θ) = 1/5
Finally, to find the angle θ, we take the inverse cosine of both sides:
θ = cos^-1(1/5)
So, the launch angle θ is the inverse cosine of 1/5.
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