A rocket is launched with a speed of 282 m/s at an angle of 41.2 degrees above the horizontal. How far does the rocket travel before it hits the ground?

Question

A rocket is launched with a speed of 282 m/s at an angle of 41.2 degrees above the horizontal. How far does the rocket travel before it hits the ground?

Answer 1

To find the horizontal distance traveled by the rocket before hitting the ground, we need to analyze the projectile motion.

Step 1: Decompose the initial velocity vector

The initial velocity can be decomposed into its horizontal and vertical components:

Vx = V * cos(θ)
Vy = V * sin(θ)

Where:
V = initial speed = 282 m/s
θ = launch angle = 41.2 degrees

Step 2: Calculate the time of flight

The time it takes for the rocket to hit the ground (time of flight) can be calculated using the vertical component of the motion. We assume the vertical displacement is zero because the rocket starts and ends at the same vertical position.

Using the kinematic equation: 𝑦 = 𝑣𝑖𝑛𝑖𝑡 + ((1/2)𝑎𝑡²), where 𝑦 = 0, 𝑣𝑖𝑛𝑖 = 𝑉𝑦 = 𝑉 sin(𝜃), and 𝑎 = -9.8 m/s² (acceleration due to gravity), we can solve for t:

0 = 𝑉 sin(𝜃)𝑡 + (1/2)(-9.8)𝑡²

Rearranging and factoring:
-4.9𝑡² + 𝑉 sin(𝜃)𝑡 = 0

Simplifying:
t (-4.9𝑡 + 𝑉 sin(𝜃)) = 0

Since time cannot be negative, we have two possible solutions: t = 0 (the initial time) and 𝑡 = (𝑉 sin(𝜃))/4.9

Therefore, the rocket will take 𝑡 = (𝑉 sin(𝜃))/4.9 seconds to hit the ground.

Step 3: Calculate the horizontal distance

To find the horizontal distance traveled, we can multiply the horizontal component of velocity (Vx) by the time of flight (t).

Horizontal distance = Vx * t
Horizontal distance = (V * cos(θ)) * ((V * sin(θ))/4.9)

Substituting the given values:
Horizontal distance = (282 * cos(41.2)) * ((282 * sin(41.2))/4.9)

Calculating this expression will give you the horizontal distance traveled by the rocket before hitting the ground.