2. A bullet is fired from the ground at an angle of 45o above the horizontal. What initial speed vo must the bullet have in order to hit a point 550 ft high on a tower located 600 ft away (ignoring air resistance)?
3. The Giant Ferris Wheel in Vienna has radius R = 30 m (Figure 1). Assuming that a time t = t0, the wheel rotates counterclockwise with a speed of 15 m/min, and is slowing at a rate of 10 m/min2. Find acceleration vector for a person seated in a car at the lowest point of the wheel.
2. To find the initial speed of the bullet, we can break down the motion into horizontal and vertical components.
First, let's focus on the vertical motion. The bullet is fired at an angle of 45 degrees above the horizontal, so the initial vertical velocity can be found using the formula:
v0y = vo * sin(theta)
where v0y is the vertical component of the initial velocity, vo is the initial velocity, and theta is the angle of 45 degrees.
Next, we can analyze the vertical motion of the bullet. The vertical displacement of the bullet is given as 550 ft, and the acceleration due to gravity is -32.2 ft/s^2 (assuming the positive direction is upwards).
Using the kinematic equation for vertical motion:
y = y0 + v0yt - (1/2)gt^2
where y is the vertical displacement, y0 is the initial vertical position (0 ft in this case), v0y is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time.
Plugging in the given values, we have:
550 = 0 + (v0 * sin(45)) * t - (1/2) * (-32.2) * t^2
Simplifying the equation, we get:
550 = (v0 / √2) * t + 16.1 * t^2
Now, let's look at the horizontal motion. The horizontal displacement of the bullet is given as 600 ft, and there is no horizontal acceleration.
Using the formula for horizontal displacement:
x = x0 + v0x * t
where x is the horizontal displacement, x0 is the initial horizontal position (0 ft in this case), v0x is the horizontal component of the initial velocity, and t is the time.
Plugging in the given values, we have:
600 = 0 + (v0 * cos(45)) * t
Simplifying the equation, we get:
600 = (v0 / √2) * t
Now, we can solve these two equations simultaneously to find the initial speed vo. We can divide the equation for the vertical motion by the equation for the horizontal motion:
550 / 600 = [(v0 / √2) * t + 16.1 * t^2] / [(v0 / √2) * t]
Simplifying the equation, we have:
550 / 600 = 1 + 16.1 * t
Now, solve for t:
550 / 600 - 1 = 16.1 * t
t ≈ 0.046 s
Next, substitute the value of t back into one of the equations to solve for vo:
600 = (v0 / √2) * 0.046
v0 ≈ 603.8 ft/s
Therefore, the initial speed vo of the bullet must be approximately 603.8 ft/s in order to hit a point 550 ft high on a tower located 600 ft away.
3. To find the acceleration vector for a person seated in a car at the lowest point of the Giant Ferris Wheel in Vienna, we need to consider both the tangential acceleration and centripetal acceleration.
First, let's determine the tangential acceleration. We are given that the wheel rotates counterclockwise with a speed of 15 m/min and is slowing at a rate of 10 m/min^2. The tangential acceleration is the rate of change of the tangential velocity.
Using the formula for tangential acceleration:
at = dv / dt
where at is the tangential acceleration, dv is the change in tangential velocity, and dt is the change in time.
Since the speed is decreasing at a rate of 10 m/min^2, the change in velocity is equal to -10 m/min^2. Considering the time t0, the change in time dt is 0.
Therefore, the tangential acceleration at the lowest point is:
at = -10 m/min^2
Next, let's determine the centripetal acceleration. The centripetal acceleration is directed towards the center of the circle and is given by the formula:
ac = v^2 / R
where ac is the centripetal acceleration, v is the tangential velocity, and R is the radius of the Ferris Wheel.
The tangential velocity is given as 15 m/min. We can convert this to m/s by dividing by 60 since there are 60 seconds in a minute.
v = 15 m/min / 60 = 0.25 m/s
Using the given radius R = 30 m, we can calculate the centripetal acceleration:
ac = (0.25 m/s)^2 / 30 m = 0.002083 m/s^2
Since the person is seated at the lowest point of the wheel, the centripetal acceleration will be directed upward and opposite to the gravitational acceleration, with a magnitude of 0.002083 m/s^2.
Therefore, the acceleration vector for a person seated in a car at the lowest point of the wheel is approximately in the upward direction with a magnitude of 0.002083 m/s^2.