Answer to Question #135746 in Physics for Sam Far

Question

Here is the given equation to utilize for all of the problems below: Vf^2=Vi^2+2aΔy.

If the girl is launched to a maximum height of 60 meters (at which her velocity = 0 m/s), and the acceleration due to gravity is -9.81 m/s, what was her initial velocity (as she left the seesaw)?

If the girl accelerated to what was the initial velocity in the last question (final velocity for this question), and she started at an initial velocity = 0 m/s at the bottom of the seesaw, what was her average acceleration (assuming that she accelerated over 1 meter of distance)?

My physics teacher had delineated to me the answer of 588 m/s^2. As I am not quite assured how to utilize the given equation (above the first question) in the context of the problem, I would greatly appreciate it if you can please demonstrate how to solve this problem with the equation, showing each step accordingly to acquire the answer of 588 m/s^2.

Answer 1

First problem

Substitute 0 for the final velocity, -9.81 for a and 60 for Δy. we use negative 9.81 because the acceleration due to gravity decelerates the girl: the acceleration due to gravity vector looks down, while the initial velocity vector look upward.

$$0=v_i^2-2\cdot9.81\cdot60,\\ v_i^2=2\cdot9.81\cdot60,\\ v_i=\sqrt{2\cdot9.81\cdot60},\\ v_i=34.3\text{ m/s}.$$

Second problem

Now substitute 1 meter for the height change Δy and 34.3 for the final velocity:

$$34.3^2=0+2a\cdot1,\\ 34.3^2=2a,\\ a=\frac{34.3^2}{2},\\ a=588\text{ m/s}^2.$$

While solving this problem, be careful with the math: remember how to use roots properly and how to solve equation. When you plug all known values into equations, think of undefined values as of x in regular equations.