A stone thrown off a bridge 20 m above a river has an initial velocity of 12 m/s at an angle of 450 above the horizontal.
a. What is the range of the stone?
b. At what velocity does the stone strike the water?
(magnitude and direction)
break the initial velocity into horizontal and vertical components.
In the vertical:
-20=0+Vvi*t-4.9t^2
solve for t.
Range: Vhi*t
To find the range of the stone, we need to calculate the horizontal distance traveled by the stone before hitting the water.
Step 1: Resolve the initial velocity into horizontal and vertical components.
The horizontal component of the initial velocity can be found using the equation:
Vx = V * cos(theta)
where
Vx = horizontal component of velocity,
V = initial velocity (12 m/s), and
theta = angle above the horizontal (45 degrees).
So, Vx = 12 m/s * cos(45 degrees) = 12 m/s * √2 / 2 = 6√2 m/s
Step 2: Calculate the time of flight.
The time of flight is the time taken by the stone to reach the water. To calculate this, we can use the vertical motion equation:
y = V * sin(theta) * t - (1/2) * g * t^2
where
y = vertical displacement (height of the bridge = 20 m),
V = initial velocity (12 m/s),
theta = angle above the horizontal (45 degrees),
t = time of flight, and
g = acceleration due to gravity (-9.8 m/s^2).
Plugging in the values:
20 m = 12 m/s * sin(45 degrees) * t - (1/2) * 9.8 m/s^2 * t^2
20 = 12 * √2 / 2 * t - 4.9 * t^2
Rearranging the equation:
4.9 * t^2 - 12 * √2 / 2 * t + 20 = 0
Solving this quadratic equation, we get two possible values for t: t1 and t2.
Step 3: Calculate the range.
The range can be found using the horizontal motion equation:
x = Vx * t
where
x = range,
Vx = horizontal component of velocity (6√2 m/s), and
t = time of flight.
Plugging in the values:
Range = 6√2 m/s * t
After calculating the time of flight (t), we can substitute it into the equation above to find the range.
Step 4: Find the velocity at impact.
The velocity at which the stone strikes the water can be found by combining the horizontal and vertical components of velocity at that moment. We can use the equations:
Vx = V * cos(theta)
Vy = V * sin(theta)
where
Vx = horizontal component of velocity,
Vy = vertical component of velocity,
V = initial velocity (12 m/s), and
theta = angle above the horizontal (45 degrees).
The magnitude of the final velocity can be found using the Pythagorean theorem:
Vf = sqrt(Vx^2 + Vy^2)
Step 5: Determine the direction of the final velocity.
The direction of the final velocity can be found using the tangent function:
tan(theta_f) = Vy / Vx
where
theta_f = angle of the final velocity with respect to the horizontal.
So, let's calculate these values:
a. Range:
1. Calculate the time of flight:
Using the quadratic formula, we get:
t = (-(-12 * √2 / 2) ± sqrt((-12 * √2 / 2)^2 - 4 * 4.9 * 20)) / (2 * 4.9)
Simplifying the equation:
t = (6√2 ± sqrt(72 - 392)) / 9.8
t ≈ (6√2 ± sqrt(-320)) / 9.8
Since the square root of a negative number is not possible in this context, we can ignore the negative solution.
t ≈ (6√2 + sqrt(-320)) / 9.8
2. Calculate the range:
Range = 6√2 m/s * t
b. Velocity at impact:
1. Calculate Vx:
Vx = 12 m/s * cos(45 degrees) = 6√2 m/s
2. Calculate Vy:
Vy = 12 m/s * sin(45 degrees) = 6√2 m/s
3. Calculate the magnitude of the final velocity:
Vf = sqrt(Vx^2 + Vy^2)
4. Calculate the direction of the final velocity:
tan(theta_f) = Vy / Vx
Please note that the calculations may involve approximations due to rounding.
To find the range of the stone, we can break down the initial velocity into its horizontal and vertical components.
The horizontal component of the initial velocity can be found using the formula:
Vx = V * cos(θ)
Where Vx is the horizontal component of the velocity, V is the initial velocity, and θ is the launch angle.
So in this case, the horizontal component of the velocity is:
Vx = 12 m/s * cos(45°)
Vx = 12 m/s * 0.7071
Vx ≈ 8.49 m/s
The time taken for the stone to hit the water can be found using the vertical component of the initial velocity and the acceleration due to gravity.
The vertical component of the velocity can be found using the formula:
Vy = V * sin(θ)
Where Vy is the vertical component of the velocity, V is the initial velocity, and θ is the launch angle.
So in this case, the vertical component of the velocity is:
Vy = 12 m/s * sin(45°)
Vy = 12 m/s * 0.7071
Vy ≈ 8.49 m/s
The time taken for the stone to hit the water can be found using the formula for time in freefall:
t = (2 * Vy) / g
Where t is the time, Vy is the vertical component of the velocity, and g is the acceleration due to gravity (-9.8 m/s^2).
So in this case, the time taken for the stone to hit the water is:
t = (2 * 8.49 m/s) / 9.8 m/s^2
t ≈ 1.73 seconds
Now, we can find the range of the stone using the formula:
Range = Vx * t
Range = 8.49 m/s * 1.73 seconds
Range ≈ 14.67 meters
So, the range of the stone is approximately 14.67 meters.
To find the velocity at which the stone strikes the water, we can use the horizontal component of the velocity (Vx) and the time taken (t).
The horizontal component of the velocity does not change during the motion of the stone because there is no horizontal force acting on it. So the horizontal component of the velocity remains constant at 8.49 m/s.
Therefore, the stone strikes the water with a horizontal velocity of 8.49 m/s.
Since we want to find the magnitude and direction of the velocity, we can use the Pythagorean theorem:
Magnitude of the velocity = √(Vx^2 + Vy^2)
Magnitude of the velocity = √((8.49 m/s)^2 + (8.49 m/s)^2)
Magnitude of the velocity ≈ √(71.92 + 71.92) m/s
Magnitude of the velocity ≈ √(143.84) m/s
Magnitude of the velocity ≈ 11.99 m/s
The direction of the velocity can be found using the inverse tangent function:
Direction = tan^(-1)(Vy / Vx)
Direction = tan^(-1)((8.49 m/s) / (8.49 m/s))
Direction ≈ 45°
So, the stone strikes the water with a magnitude of approximately 11.99 m/s and at an angle of 45° above the horizontal.