Rock thrown from cliff, find initial and final velocities

[kitkat87]

Homework Statement

I'm taking grade 12 U physics and I'm having a really difficult time understanding the answers given compared to the answers I arrive at.

For example:

The question reads:

A rock was thrown horizontally from a 25.0 m high cliff and landed 15.0 m from the base of the cliff. Determine the initial speed with which the rock was thrown, as well as its final velocity.

2. The attempt at a solution

Vertical
V1y = 0.0 m/s
ay=9.8 m/s^2
Displacement = 25 m

Solve for y-component

25m = (0.0 m/s)(change in time) + 1/2(9.8 m/s^2)(change in time)^2

Rearrange

(change in time)^2 = 2(25)/9.8m/s^2
change in time = Sqrt 2(25) / 9.8m/s^2
change in time = 0.7215375318230077

This is the answer I arrived at.

My workbook however, says the answer is 2.26. Can someone please explain to me how my workbook arrived that that answer? I cannot move on in the question until I solve this.

So far every example in my course provides the correct calculations but all of the final answers are completely incorrect.

 

[BiGyElLoWhAt]

You did order of operations wrong. Work it out again from the first line and keep your parentheses.

 

[ehild]

Homework Statement

The question reads:

A rock was thrown horizontally from a 25.0 m high cliff and landed 15.0 m from the base of the cliff. Determine the initial speed with which the rock was thrown, as well as its final velocity.

2. The attempt at a solution

Vertical
V1y = 0.0 m/s
ay=9.8 m/s^2
Displacement = 25 m

Solve for y-component

25m = (0.0 m/s)(change in time) + 1/2(9.8 m/s^2)(change in time)^2

Rearrange

(change in time)^2 = 2(25)/9.8m/s^2
change in time = Sqrt 2(25) / 9.8m/s^2
change in time = 0.7215375318230077

Welcome to PF!
You miss parentheses. (change in time)^2 = 2*25/9.8 = 5.102. Take the square root to get the time : The correct formula is (change in time) = Sqrt ((2*25) / 9.8)
You took the square root of 50 and divided it by 9.8, which is wrong.

ehild

 

[kitkat87]

Thank you both so much! you're amaz-o

My entire workbook leaves the 9.8m/s^2 throughout all of the examples so I thought that it stayed that way.

You've just made my life so much easier.

 

[HalfLostAndSearching]

I can understand your frustration with the discrepancies between your calculations and the answers provided in your workbook. It is important to remember that in physics, there can be multiple ways to arrive at the correct answer and sometimes there may be slight variations in the final values due to rounding or different methods used.

In this specific problem, it appears that your workbook may be using a different approach to solving for the initial and final velocities. It is possible that they are using equations for projectile motion or conservation of energy to solve for these values. Without seeing the specific equations and calculations used in your workbook, it is difficult for me to explain how they arrived at the answer of 2.26.

However, the important thing to remember is that as long as your calculations are accurate and your approach is correct, your answer of 0.7215375318230077 is also a valid solution. I encourage you to discuss this with your teacher or classmates to understand the different methods used and how they can lead to slightly different answers.

In science, it is important to focus on the process and understanding the concepts rather than just getting the "right" answer. Keep up the good work and continue to question and seek understanding in your studies. Good luck!