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
v
at an angle θ with respect to the horizontal. Let the building be 49.0 m tall, the initial horizontal velocity be 9.00 m/s, and the initial vertical velocity be 11.5 m/s. Choose your coordinates such that the positive y-axis is upward, and 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
See Related Questions: Fri, 9-9-11,11:20
AM.
To find the ball's maximum height above the ground and the time it takes to reach that height, we can use the equations of motion for projectile motion.
In this case, we have the initial vertical velocity (Vy) as 11.5 m/s and the initial horizontal velocity (Vx) as 9.00 m/s. Since the ball is thrown off the top of a building, we can consider the building's height (h) as the initial vertical position (y0), which is 49.0 m above the ground.
We need to break down the initial velocity into its vertical and horizontal components using trigonometry. The vertical component is given by Vy = V * sin(θ), and the horizontal component is given by Vx = V * cos(θ), where V is the initial velocity and θ is the angle with respect to the horizontal.
Given that Vy = 11.5 m/s and Vx = 9.00 m/s, we can solve for V and θ using the inverse trigonometric functions.
Vy = V * sin(θ)
11.5 = V * sin(θ)
Vx = V * cos(θ)
9.00 = V * cos(θ)
To find V and θ, divide the second equation by the first equation:
9.00 / 11.5 = cos(θ) / sin(θ)
0.7826 = tan(θ)
Using the inverse tangent function, we can find θ:
θ = tan^(-1)(0.7826)
θ ≈ 39.93 degrees
Now, we can find V using Vy = V * sin(θ):
11.5 = V * sin(39.93)
V = 11.5 / sin(39.93)
V ≈ 18.13 m/s
With V and θ determined, we can proceed to find the maximum height (ymax) and the time it takes to reach that height (tmax).
The time to reach the maximum height can be found using the vertical component of velocity, Vy, and the acceleration due to gravity, g = 9.8 m/s^2. At the maximum height, the vertical component of velocity will be zero. Using the equation:
Vy = Vy0 - g * t
0 = 11.5 - 9.8 * t
Solving for t:
9.8 * t = 11.5
t = 11.5 / 9.8
t ≈ 1.17 seconds
To find the maximum height, we can use the equation:
ymax = y0 + Vy0 * t - (1/2) * g * t^2
Substituting the values:
ymax = 49.0 + 11.5 * 1.17 - (1/2) * 9.8 * (1.17)^2
Simplifying the equation will give us the value of ymax.