A projectile is fired at 32.5° above the horizontal. Its initial speed is equal to 160.0 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile?

Question

A projectile is fired at 32.5° above the horizontal. Its initial speed is equal to 160.0 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile?

Answer 1

See previous post: Tue, 12-1-15, 12:33 PM.

Answer 2

To find the maximum height reached by the projectile, we can use the concept of projectile motion and break it down into two components: horizontal and vertical.

First, let's find the time it takes for the projectile to reach its maximum height. Since there is no initial vertical velocity (since the initial velocity is only in the horizontal direction), the vertical motion of the projectile can be modeled as a vertically launched projectile. The only force acting on it is gravity, which leads to a constant acceleration in the vertical direction. The vertical component of the initial velocity can be found using trigonometry:

Vertical Component of the Initial Velocity (Vy) = Initial Speed * sin(θ)
where θ is the angle of launch.

Given:
Initial Speed (V) = 160.0 m/s
Launch Angle (θ) = 32.5°

Vy = 160.0 m/s * sin(32.5°)

Now, we can use the fact that the time it takes for the projectile to reach its maximum height is equal to the time it takes for the vertical component of the velocity (Vy) to become zero. The vertical motion can be described using the equation:

Vy = Vy0 + g * t
where
Vy is the vertical component of velocity at time t,
Vy0 is the vertical component of initial velocity,
g is the acceleration due to gravity,
t is the time.

Since the projectile reaches its maximum height when Vy = 0, we can solve for t:

0 = Vy0 + g * t
t = -Vy0 / g

Given:
g (acceleration due to gravity) ≈ 9.8 m/s^2 (assuming Earth's gravity)

Now, substitute the values in the equation:

t = -Vy0 / g = -(160.0 m/s * sin(32.5°)) / (9.8 m/s^2)

Next, we can find the maximum height (H) reached by the projectile using the equation:

H = Vy0 * t + (1/2) * g * t^2

Substitute the values into the equation:

H = (160.0 m/s * sin(32.5°)) * [-(160.0 m/s * sin(32.5°)) / (9.8 m/s^2)] + (1/2) * (9.8 m/s^2) * [-(160.0 m/s * sin(32.5°)) / (9.8 m/s^2)]^2

Simplify and calculate:

H = (160.0 m/s)^2 * (sin(32.5°))^2 / (2 * 9.8 m/s^2)

This will give you the maximum height reached by the projectile.