Rock thrown from cliff, find initial and final velocities

AI Thread Summary
A student is struggling with a physics problem involving a rock thrown horizontally from a 25.0 m high cliff, needing to determine the initial speed and final velocity. They calculated the time of fall using the formula for vertical displacement but arrived at an incorrect value. Forum members pointed out that the student misapplied the order of operations, specifically in handling parentheses during calculations. The correct approach involves using the formula (change in time) = Sqrt((2*25) / 9.8) to find the time of fall, which should yield approximately 2.26 seconds. This clarification helped the student understand their mistake and move forward with the problem.
kitkat87
Messages
6
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top