how long will it take a bullet fired from a cliff at an initial velocity of 700m per second at an angle of 30 Degree below the horizontal to reach the ground 200m below?answer
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find the horizontal and vertical components of the initial velocity.
vertical=700*sin30=350m/s
hf=hi+vi*t-4.9t^2
0=200+350*t-4.9t^2
put that quadratic in standard form
ax^2 + bx + c=0
and solve for t with the quadratic equation
To determine the time it takes for the bullet to hit the ground, we can break down the problem into horizontal and vertical components.
First, let's find the vertical component of the velocity. Given that the initial velocity of the bullet is 700 m/s and it is fired at an angle of 30 degrees below the horizontal, we can find the vertical component using the formula:
Vertical component velocity (Vy) = Initial velocity (V) * sin(theta)
Where:
V = 700 m/s (initial velocity)
theta = 30 degrees
Vy = 700 m/s * sin(30 degrees)
Vy = 700 m/s * 0.5
Vy = 350 m/s
Now, let's find the time it takes for the bullet to reach the ground. We can use the vertical displacement and vertical component of velocity to solve for time using the formula:
Vertical displacement (d) = Vertical component velocity (Vy) * time (t) + 0.5 * acceleration (a) * time (t)^2
Where:
d = -200 m (negative since it is below the initial position)
Vy = 350 m/s (vertical component velocity)
a = -9.8 m/s^2 (acceleration due to gravity)
Plugging in the values, we get:
-200 m = (350 m/s) * t + 0.5 * (-9.8 m/s^2) * t^2
Simplifying the equation:
0.5 * (-9.8 m/s^2) * t^2 + (350 m/s) * t + 200 m = 0
Now, we can solve this quadratic equation to find the value of time. However, since the bullet will hit the ground, we only need to consider the positive value of time.
Using the quadratic formula:
t = (-b + √(b^2 - 4ac))/2a
Where:
a = 0.5 * (-9.8 m/s^2)
b = 350 m/s
c = 200 m
Plugging in the values, we can calculate the time it takes for the bullet to hit the ground.
To find out how long it will take for the bullet to reach the ground, we can use the equations of motion in two dimensions.
First, let's break down the initial velocity into its horizontal and vertical components.
Given:
Initial velocity (v0) = 700 m/s
Angle with respect to the horizontal (θ) = 30 degrees below the horizontal
Vertical distance (y) = -200 m (negative because it is below the starting point)
Horizontal component of initial velocity (v0x) = v0 * cos(θ)
v0x = 700 m/s * cos(30°) = 700 m/s * sqrt(3)/2 = 700 * 0.866 = 606 m/s
Vertical component of initial velocity (v0y) = v0 * sin(θ)
v0y = 700 m/s * sin(30°) = 700 m/s * 0.5 = 350 m/s
Now, we can use the equation for vertical displacement to find the time it takes for the bullet to reach the ground:
y = v0yt + (1/2)gt^2
where:
y = vertical distance (-200 m)
v0y = vertical component of initial velocity (350 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)
Plugging in the values, we have:
-200 = (350)t + (1/2)(-9.8)t^2
Rearranging the equation to form a quadratic equation:
-4.9t^2 + 350t + 200 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where a = -4.9, b = 350, and c = 200.
Solving for t gives us two possible values. However, we are interested in the positive value since time cannot be negative:
t = (-350 + sqrt(350^2 - 4(-4.9)(200))) / (2(-4.9))
Calculating the result:
t = (-350 + sqrt(122500 + 3920)) / -9.8
t = (-350 + sqrt(126420)) / -9.8
t = (-350 + 355.22) / -9.8
t = 5.22 / -9.8
t ≈ -0.532 seconds
We ignore the negative result since time cannot be negative in this context.
Therefore, it will take approximately 0.532 seconds for the bullet to reach the ground 200 m below.