If a ball has an initial speed of 26.6 m/s and is launched at an angle of 63.2 degrees above the horizontal, what is the tallest obstacle that the ball can clear?
My goodness, if you can't begin to answer any of these seven questions, maybe you shouldn't be taking physics. If you want to pass this class, you should be doing your work yourself so that you can pass your tests.
I will be happy to critique your thinking.
To find the tallest obstacle that the ball can clear, we need to determine the projectile's maximum height. We can do this by applying the equations of motion for projectile motion.
The horizontal and vertical motions of the projectile are independent of each other. Firstly, let's analyze the vertical motion:
The initial velocity in the vertical direction (Vy) can be found by decomposing the initial velocity (26.6 m/s) into its vertical and horizontal components:
Vy = V * sin(θ)
= 26.6 m/s * sin(63.2°).
Now, we can use the vertical motion equation to find the maximum height (H) reached by the projectile:
H = (V^2 * sin^2(θ)) / (2 * g),
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the known values:
H = (26.6 m/s)^2 * sin^2(63.2°) / (2 * 9.8 m/s^2).
Calculating this expression will give us the maximum height reached by the projectile.
To find the tallest obstacle cleared, we need to consider the difference between the maximum height (H) and the initial height of the projectile. If the obstacle's height is less than this difference, the ball can clear it. The initial height should be provided to determine the exact answer.