To determine the time it takes for the body to reach the ground, we can analyze the vertical motion separately. We'll use the following kinematic equation:
y = v₀t + (1/2)at²
where: y = vertical displacement (which will be -h, negative since we're measuring downward) v₀ = initial vertical velocity t = time a = acceleration due to gravity (approximately -9.8 m/s², negative since it acts downward)
Let's calculate the time it takes for the body to reach the ground:
Since the body is projected at an angle of 30° with the horizontal, the initial vertical velocity can be determined as:
v₀y = v₀ * sin(θ) = 49 m/s * sin(30°) = 49 m/s * 0.5 = 24.5 m/s
Now, using the kinematic equation, we can solve for time (t) when y = -h (height of the body at the ground):
-h = v₀yt + (1/2)at²
Since the body starts at ground level (h = 0), the equation simplifies to:
0 = v₀yt + (1/2)at²
0 = (24.5 m/s)t + (1/2)(-9.8 m/s²)t²
0 = 24.5t - 4.9t²
Rearranging the equation:
4.9t² - 24.5t = 0
Factorizing t:
t(4.9t - 24.5) = 0
This equation has two solutions: t = 0 (which is the initial time) and 4.9t - 24.5 = 0.
Solving 4.9t - 24.5 = 0 for t:
4.9t = 24.5 t = 24.5 / 4.9 t ≈ 5 seconds
Therefore, it will take approximately 5 seconds for the body to reach the ground.
To calculate the horizontal distance traveled by the body (range), we can use the equation:
x = v₀x * t
where: x = horizontal distance traveled v₀x = initial horizontal velocity t = time (which is 5 seconds)
The initial horizontal velocity (v₀x) can be calculated as:
v₀x = v₀ * cos(θ) = 49 m/s * cos(30°) = 49 m/s * (√3/2) = 49 m/s * 0.866 ≈ 42.434 m/s
Substituting the values into the equation:
x = (42.434 m/s) * 5 s x ≈ 212.17 m
Therefore, the body will strike the ground approximately 212.17 meters away from the starting point.