Rock's Velocity at Bottom of Hole

AI Thread Summary
The discussion revolves around calculating the velocity of a rock as it hits the bottom of a 10 m deep hole after being tossed upwards at 20 m/s. The user calculated the final velocity to be -24 m/s, interpreting downward motion as negative, while the textbook states the answer is +24 m/s, indicating speed without direction. The confusion arises from the distinction between speed (magnitude) and velocity (directional). The user correctly identifies that the rock's final velocity should be negative if considering direction, but the textbook likely refers to the speed, which is always positive. The conversation highlights the importance of understanding the difference between speed and velocity in physics problems.
sona1177
Messages
171
Reaction score
1

Homework Statement



A rock is tossed straight up with a speed of 20 m/s. When it returns, it falls into a hole 10 m deep.

What is the rock's velocity as it hits the bottom of the hole?



Homework Equations


V = Vo + at
Y - Yo = Vot + .5at2
v2 = vo2 + 2a(Y - Yo)
Y - Yo = .5(Vo + V)t

The Attempt at a Solution


My answer is -24 m/s but the book reports positive 24 m/s. I don't see why this is the answer if the rock is moving downward. I've identified up as the positive direction. I got my answer by using equation v2 = vo2 + 2a(Y - Yo) where V was unknown, Vo =20 m/s, a=-9.8, displacement=-10. I know the calculator only gives positive square roots but you can have negative square roots also and since this object is moving down, shouldn't the velocity be -24 m/s. The question later asks for time and according to the book, the value is 4.5 seconds using equation Y - Yo = Vot + .5at2 but this only true if you plug is -24 m/s not simply positive 24 m/s. This is why I don't see why the answer book doesn't report -24 m/s as the final velocity before it hits the bottom of the whole. This is very frustrating.
 
Physics news on Phys.org
You are right. If the initial velocity is positive, the final one is negative. 24 m/s is the final speed.

ehild
 
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 '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