A mortar shell is launched with a velocity of 100m/s at an angle of 30 degree above the horizontal from a point on a cliff 50m above a level plain below. How far from the base of the cliff does the mortar shell strike the ground?
The distance from the base of the cliff to where the mortar shell strikes the ground is approximately 122.5 meters.
To find the horizontal distance traveled by the mortar shell, we need to determine the time it takes to reach the ground. We can break down the given information as follows:
Initial horizontal velocity (Vx) = initial velocity (100 m/s) * cosine(angle)
= 100 m/s * cos(30°)
= 100 m/s * 0.866
≈ 86.6 m/s
Using the equation of motion for vertical displacement, considering the initial vertical velocity (Vy) as 0 and the displacement (dy) as -50 m (since it is going downwards):
dy = Vy * t + (1/2) * g * t^2, where g is the acceleration due to gravity (9.8 m/s^2)
Substituting the values, we have:
-50 = 0 * t + (1/2) * 9.8 * t^2
-50 = 4.9 * t^2
Simplifying the equation:
t^2 = -50 / 4.9
t^2 ≈ -10.2
Since time cannot be negative, we can ignore the negative sign and take the square root of the equation:
t ≈ √10.2
t ≈ 3.19 s
Now that we have the time taken to reach the ground, we can find the horizontal distance traveled:
displacement (dx) = Vx * t
= 86.6 m/s * 3.19 s
≈ 276.09 m
Therefore, the mortar shell strikes the ground approximately 276.09 meters from the base of the cliff.
To find the horizontal distance traveled by the mortar shell before it strikes the ground, we need to calculate the projectile's horizontal range.
First, let's break down the initial velocity of the shell into its horizontal and vertical components. The horizontal component (Vx) can be calculated using the equation:
Vx = v * cos(theta)
Where v is the initial velocity (100 m/s) and theta is the angle (30 degrees).
Vx = 100 m/s * cos(30 degrees)
Vx = 100 m/s * 0.866
Vx = 86.6 m/s
The vertical component (Vy) can be calculated using the equation:
Vy = v * sin(theta)
Where v is the initial velocity (100 m/s) and theta is the angle (30 degrees).
Vy = 100 m/s * sin(30 degrees)
Vy = 100 m/s * 0.5
Vy = 50 m/s
Now, let's find the time it takes for the shell to reach the ground. We can use the equation:
time = (2 * Vy) / g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
time = (2 * 50 m/s) / 9.8 m/s^2
time = 10.20 s
Next, we can calculate the horizontal range (distance) traveled by the shell using the equation:
distance = Vx * time
distance = 86.6 m/s * 10.20 s
distance = 884.52 m
Therefore, the mortar shell strikes the ground approximately 884.52 meters away from the base of the cliff.