A. You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity at an angle θ with respect to the horizontal. Let the building be 50.0 m tall, the initial horizontal velocity be 9.00 m/s, and the initial vertical velocity be 12.0 m/s. Choose your coordinates such that the positive y-axis is upward, the x-axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, find the ball’s maximum height above the ground, and the time it takes to reach the maximum height. (b) Repeat your calculations choosing the origin at the base of the building.
a. Vf^2 = Vo^2 + 2gd,
d = 50 + (Vf^2 - Vo^2) / 2g,
d = 50 + (0 - (12)^2) / -19.6,
d = 50 + 7.35 = 57.35m above ground.
Vf = V0 + gt,
t = (Vf - Vo) / g,
t(up) = (0 - 12) / -9.8 = 1.22s.
b. d = (Vf^2 - Vo^2) / 2g,
d = (0 - (12)^2) / -19.6 = 7.35m above
ground.
t(up) = (Vf - Vo) / g,
t(up) = (0 - 12) / -9.8 = 1.22s.
good answer but part one of b does not give much accuracy
(a) To find the maximum height above the ground and the time it takes to reach that height, we can use the equations of motion for projectile motion.
Given:
Initial horizontal velocity, Vx = 9.00 m/s
Initial vertical velocity, Vy = 12.0 m/s
Height of the building, h = 50.0 m
1. Maximum height (y-coordinate):
The maximum height occurs when the vertical component of velocity (Vy) becomes zero.
Using the equation: V^2 = u^2 + 2as, where V = 0, u = Vy, and a = -g (acceleration due to gravity, taken as -9.8 m/s^2),
0 = (12.0)^2 + 2(-9.8)(y_max)
144.0 = -19.6y_max
y_max = -144.0 / 19.6
y_max ≈ -7.35 m
Since we have chosen the positive y-axis upward, the maximum height is above the ground, so we take the absolute value:
Maximum height = |y_max| ≈ 7.35 m
2. Time to reach the maximum height:
Using the equation: v = u + at, where v = 0, u = Vy, and a = -g,
0 = 12.0 - 9.8t_max
t_max = 12.0 / 9.8
t_max ≈ 1.22 s
Therefore, the ball reaches the maximum height of approximately 7.35 m above the ground after approximately 1.22 seconds.
(b) Let's repeat the calculations with the origin at the base of the building.
1. Maximum height (y-coordinate):
Since the building is at the origin now, the maximum height is simply the difference between the ball's initial height and the height of the building.
Maximum height = 0 - (-50.0)
Maximum height = 50.0 m
2. Time to reach the maximum height:
Using the same calculations as in part (a), the time to reach the maximum height remains the same:
t_max = 1.22 s
Therefore, when choosing the origin at the base of the building, the ball reaches a maximum height of 50.0 m above the ground, and it takes approximately 1.22 seconds to reach that height.
To solve this projectile motion problem, we can use the equations of motion and the concepts of kinematics. I'll guide you through the process step-by-step.
(a) Coordinate System with the Origin at the Point of Release:
1. Break down the initial velocity into its horizontal and vertical components.
- The initial horizontal velocity (Vx) is 9.00 m/s.
- The initial vertical velocity (Vy) is 12.0 m/s.
2. Determine the time it takes to reach the maximum height (t_max).
- The vertical motion can be treated as free-fall with an initial velocity of Vy.
- Use the equation: Vy = V0y + gt, where g is the acceleration due to gravity (-9.8 m/s^2).
- We know that the final vertical velocity at the maximum height is zero, so V0y = -12.0 m/s (negative due to the upward direction).
- Rearrange the equation to solve for t: t = (Vf - V0y) / g.
- Plugging in the values: t_max = (0 - (-12.0)) / (-9.8) = 1.224 s.
3. Find the maximum height above the ground (h_max).
- The vertical motion can be described by the equation: h = V0y * t + (1/2) * g * t^2.
- In this case, we want to find h_max, so we substitute t with t_max.
- Simplify the equation: h_max = -12.0 * 1.224 + (1/2) * (-9.8) * (1.224)^2 = 7.57 m.
So, with the chosen coordinate system, the ball reaches a maximum height of 7.57 meters above the ground after 1.224 seconds.
(b) Coordinate System with the Origin at the Base of the Building:
1. Adjust the values for the problem.
- The building's height is now 0 m since we shifted the origin to the base.
- The initial vertical velocity (Vy) remains the same at 12.0 m/s.
2. Determine the time it takes to reach the maximum height (t_max).
- The approach remains the same as in (a), using the equation: Vy = V0y + gt.
- Plugging in the values: t_max = (0 - 12.0) / (-9.8) = 1.224 s.
3. Find the maximum height above the ground (h_max).
- The height in this case is already 0 m.
- So, h_max = 0 m.
With the origin at the base of the building, the ball does not move vertically, and so the maximum height is 0 meters above the ground. The time to reach the maximum height remains the same at 1.224 seconds.
By comparing the two scenarios, you can observe how the choice of coordinate system affects the interpretation of the problem and the resulting calculations.