A rifle bullet is fired at an angle of 28.3° below the horizontal with an initial velocity of 177 m/s from the top of a cliff 52.2 m high. How far from the base of the cliff does it strike the level ground below?
Horizonal velocity: 177 cos28.3
vertical velocity: - 177sin28.3
time to hit ground.
-52.2=-177sin28.3*t-1/2 g t^2 solve that for time t.
horizonal distance=horizonalVelocity*t
To solve this problem, we can break down the initial velocity of the bullet into its horizontal and vertical components.
Given information:
Initial velocity (v) = 177 m/s
Angle (θ) = 28.3°
Height of the cliff (h) = 52.2 m
Step 1: Finding the vertical component of the velocity (Vy)
The vertical component of the velocity (Vy) can be found using the formula:
Vy = v * sin(θ)
Vy = 177 m/s * sin(28.3°)
Vy = 177 m/s * 0.4671
Vy ≈ 82.6527 m/s
Step 2: Finding the time taken to reach the ground (t)
The time taken for the bullet to reach the ground can be calculated using the vertical component of velocity (Vy) and the formula for vertical motion:
h = Vy * t + (1/2) * g * t^2
Rearranging the equation, we get:
t = sqrt((2 * h) / g)
Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Plugging in the values, we have:
t = sqrt((2 * 52.2 m) / 9.8 m/s^2)
t ≈ sqrt(10.64)
t ≈ 3.26 s
Step 3: Finding the horizontal distance (dx)
The horizontal distance traveled by the bullet can be found using the horizontal component of the velocity (Vx) and the formula for horizontal motion:
dx = Vx * t
Since there is no acceleration horizontally, the horizontal component of the velocity (Vx) is equal to the initial velocity (v) multiplied by cosine of the angle (θ):
Vx = v * cos(θ)
Plugging in the values, we have:
Vx = 177 m/s * cos(28.3°)
Vx = 158.886 m/s
dx = Vx * t
dx = 158.886 m/s * 3.26 s
dx ≈ 517.81 m
Therefore, the bullet strikes the level ground approximately 517.81 meters from the base of the cliff.
To solve this problem, we can break it down into two components: the horizontal and vertical motions of the bullet.
First, let's analyze the vertical motion. We can use the equation of motion to find the time it takes for the bullet to reach the ground:
h = vit + (1/2)gt^2
Where:
- h is the height (52.2 m)
- vi is the initial vertical velocity (unknown but can be calculated from the initial velocity and the angle of 28.3°)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time
Since the bullet is fired at an angle below the horizontal, the initial vertical velocity can be calculated as:
vi = 177 m/s * sin(28.3°)
Now we can substitute these values into the equation to solve for t:
52.2 = (177 * sin(28.3°))t - (1/2)(9.8)t^2
Rearranging the equation to put it in standard quadratic form:
-4.9t^2 + (177 * sin(28.3°))t - 52.2 = 0
Solving this quadratic equation will give us the time it takes for the bullet to hit the ground. We can use the quadratic formula or factoring methods to find the roots. Let's use the quadratic formula:
t = [-b ± √(b^2 - 4ac)] / (2a)
Plugging in the values of a, b, and c:
t = [-(177 * sin(28.3°)) ± √((177 * sin(28.3°))^2 - 4(-4.9)(-52.2))] / (2 * (-4.9))
Now we have two possible values of t. However, the negative value does not make sense in this context since time cannot be negative. So, we will only consider the positive value.
Next, let's analyze the horizontal motion. We can use the equation of motion for horizontal motion to find the horizontal distance traveled by the bullet:
d = vix * t
Where:
- d is the horizontal distance
- vix is the initial horizontal velocity (unknown but can be calculated from the initial velocity and the angle of 28.3°)
- t is the time we calculated earlier
The initial horizontal velocity can be calculated as:
vix = 177 m/s * cos(28.3°)
Now we can substitute these values into the equation to solve for d:
d = (177 * cos(28.3°)) * t
Calculating the value of d will give us the horizontal distance from the base of the cliff where the bullet strikes the level ground below.