Question

If a projectile has an initial velocity of 20m/s and was projected to an angle of 45°, what is the maximum height reached by the projectile?

Answer 1

To determine the maximum height reached by the projectile, we can analyze the motion along the vertical direction separately. Given the initial velocity (v₀) and launch angle (θ), we can calculate the maximum height (h_max) using the following steps:

1. Resolve the initial velocity into its vertical and horizontal components: Vertical component (v₀y) = v₀ * sin(θ) Horizontal component (v₀x) = v₀ * cos(θ)

2. Calculate the time it takes for the projectile to reach its highest point. The time of flight (t_flight) to reach the highest point can be determined using the equation: t_flight = (2 * v₀y) / g where g is the acceleration due to gravity (approximately 9.8 m/s²).

3. Determine the maximum height (h_max) using the formula: h_max = (v₀y²) / (2 * g)

Now, let's calculate the maximum height reached by the projectile:

Given: v₀ = 20 m/s θ = 45° g = 9.8 m/s²

1. Resolving the initial velocity: v₀y = v₀ * sin(θ) = 20 * sin(45°) ≈ 14.14 m/s (vertical component) v₀x = v₀ * cos(θ) = 20 * cos(45°) ≈ 14.14 m/s (horizontal component)

2. Calculating the time of flight: t_flight = (2 * v₀y) / g = (2 * 14.14) / 9.8 ≈ 2.88 s

3. Determining the maximum height: h_max = (v₀y²) / (2 * g) = (14.14²) / (2 * 9.8) ≈ 10.20 m

Therefore, the maximum height reached by the projectile is approximately 10.20 meters.