To determine the velocity of a projectile after a certain time, we need to consider its horizontal and vertical components separately.
Given: Initial velocity (u) = 200 m/s Launch angle (θ) = 30 degrees Time (t) = 16 seconds
First, let's calculate the horizontal and vertical components of the initial velocity:
Horizontal component (u_x) = u * cos(θ) Vertical component (u_y) = u * sin(θ)
Using trigonometric functions, we can find their values:
u_x = 200 m/s * cos(30°) u_y = 200 m/s * sin(30°)
u_x ≈ 200 m/s * √3/2 ≈ 200 * 1.732/2 ≈ 173.2 m/s u_y ≈ 200 m/s * 1/2 ≈ 100 m/s
Next, we can calculate the vertical displacement (s_y) using the formula:
s_y = u_y * t + (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s²).
s_y = 100 m/s * 16 s + (1/2) * 9.8 m/s² * (16 s)^2 s_y = 1600 m + 7.84 m/s² * 256 s² s_y = 1600 m + 2007.04 m s_y ≈ 3607.04 m
Since the projectile was launched from the ground, the vertical displacement represents the height reached.
Now, let's calculate the horizontal displacement (s_x):
s_x = u_x * t s_x = 173.2 m/s * 16 s s_x ≈ 2771.2 m
Finally, we can calculate the resultant velocity (v) using the Pythagorean theorem:
v = √(s_x² + s_y²) v = √(2771.2 m)² + (3607.04 m)² v ≈ √(7671843.44 m² + 13038880.1216 m²) v ≈ √20710723.5616 m² v ≈ 4549.88 m/s
Therefore, the velocity of the projectile after 16 seconds would be approximately 4549.88 m/s.