Question

A projectile with an initial velocity of 256.8 m/s was observed to make a 55° angle above the horizontal and traveled it's whole trajectory in 50 seconds. How far did it travel along the horizontal axis? Note: Use 4 decimal places?

Answer 1

To determine the horizontal distance traveled by the projectile, we can use the horizontal component of the initial velocity and the time of flight.

Given: Initial velocity (v0) = 256.8 m/s Launch angle (θ) = 55° Time of flight (t) = 50 seconds

First, we need to find the horizontal component of the initial velocity, which can be calculated using the equation:

v₀x = v₀ * cos(θ)

where v₀x is the horizontal component of the initial velocity.

Plugging in the values, we have:

v₀x = 256.8 m/s * cos(55°)

Next, we can calculate the horizontal distance traveled using the equation:

d = v₀x * t

Plugging in the values, we have:

d = (256.8 m/s * cos(55°)) * 50 s

Calculating this expression using a calculator or computer, we find:

d ≈ 11519.5392 meters

Rounded to four decimal places, the horizontal distance traveled by the projectile is approximately 11519.5392 meters.