Answer:
The shell is in the air for approximately 10.2041 seconds,
which is closest to option C: 10.2 s.
Step-by-step explanation:
To find the time the shell is in the air, you can use the kinematic equation for vertical motion. The horizontal motion and vertical motion are independent of each other, so we can focus on the vertical motion to determine the time of flight.
The initial velocity of the shell can be divided into its horizontal and vertical components.
The vertical component can be calculated as:
Vertical velocity (Vy) = initial velocity (V) * sin(angle)
= 100 m/s * sin(30°)
≈ 50 m/s
Now, we can use the equation for vertical motion:
Vertical displacement (s) = (initial vertical velocity * time) + (0.5 * acceleration * time^2)
Since the shell is launched vertically upwards (against gravity), the acceleration is negative due to gravity (approximately -9.8 m/s²). The final vertical displacement is zero (the shell returns to the ground level).
0 = (50 m/s * t) + (0.5 * (-9.8 m/s²) * t^2)
Simplify the equation:
-4.9 t^2 + 50 t = 0
Factor out t:
t * (-4.9 t + 50) = 0
This gives us two possible solutions:
t = 0 (initial time)
-4.9 t + 50 = 0
-4.9 t = -50
t ≈ 10.2041 seconds
Since the negative time doesn't make physical sense in this context, we discard it.
So, the shell is in the air for approximately 10.2041 seconds, which is closest to option C: 10.2 s.