Let's break down the questions and the explanations given by "drwls" in the post.
For question #2, drwls used the formula "8.0 m = (Vo^2 / g) * sin(2A)" to solve for the takeoff speed (Vo).
In this formula:
- "Vo" represents the takeoff speed, which is what we are trying to find.
- "g" represents the acceleration due to gravity.
- "A" represents the angle at which the long jumper leaves the ground, which is 45 degrees in this case.
- The sin(2A) term accounts for the fact that the jumper's motion can be split into a vertical component and a horizontal component, with the takeoff angle being double the angle measured from the horizontal.
Now, let's move on to question #1. To find the horizontal distance traveled before the ball lands on the building, drwls used the value of "27/sqrt2" for both the horizontal velocity (Vx) and vertical velocity (Voy) components.
In this case:
- "27" represents the initial speed of the ball.
- "sqrt2" is the square root of 2, which appears because the ball is launched at a 45-degree angle, and the horizontal and vertical velocities are equal.
To find the time of flight (t), drwls used the equation "Y = 1 + Voy * t - (g/2) * t^2 = 13" and solved for "t". "Y" represents the initial height of the ball, which is 1 meter above the ground, and "13" is the height of the roof.
To find the horizontal distance (X), drwls used the equation "X = Vx * t", which simply represents the distance traveled in the horizontal direction during the time "t" obtained earlier.
Finally, for question #3, drwls suggested that you use what you have learned from the previous questions and try to solve it yourself. However, you mentioned that you still need help with this question. Could you provide more details about what specifically you are struggling with in question #3?