Rock thrown from cliff, find initial and final velocities

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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.