Rock thrown from cliff, find initial and final velocities

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SUMMARY

The discussion focuses on solving a physics problem involving a rock thrown horizontally from a 25.0 m high cliff, landing 15.0 m away. The user initially calculated the time of flight incorrectly due to improper use of parentheses in the formula. The correct calculation shows that the time of flight is approximately 2.26 seconds, derived from the equation (change in time) = Sqrt((2 * 25) / 9.8). The final velocity can be determined using this time and the horizontal distance traveled.

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kitkat87
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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.
 
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You did order of operations wrong. Work it out again from the first line and keep your parentheses.
 
kitkat87 said:

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
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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