How Fast Must the Rock Be Thrown to Reach the Naturalist?

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Homework Help Overview

The problem involves a scenario where a naturalist is being lifted by a hoist while her friend attempts to throw a rock to her. The challenge is to determine the minimum initial speed required for the rock to reach the naturalist, who is positioned 2.50 m above the friend. The context includes gravitational acceleration and neglects air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using kinematic equations to relate the motion of the rock and the naturalist. Some suggest employing conservation of energy to equate the kinetic energy of the rock with the total energy of the naturalist at her height. Others question the conditions for the rock's motion, particularly regarding its velocity at the apex of its flight.

Discussion Status

The discussion is active, with participants exploring different methods to approach the problem. Some have shared their previous experiences with similar problems, indicating a variety of perspectives on how to tackle the question. There is no explicit consensus yet on the best approach.

Contextual Notes

Participants are considering the implications of the rock's initial velocity and the timing of its ascent relative to the naturalist's height and speed. The problem's constraints, such as the assumption of negligible air resistance and the specific values given for gravitational acceleration, are acknowledged but not resolved.

kalupahana
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Free fall, problem, please help

Homework Statement


A hoist is lifting a naturalist to the top of a cliff at 2.03 m/s vertically. The naturalist suddenly realizes she has left her pet rock behind. A friend picks it up and tosses it straight upward. If the naturalist is 2.50 m above her friend, what is the minimum initial speed the pet rock must have to reach the naturalist?
(consider gravitational acceleration as 10ms-2 & air resistance negligible)

Homework Equations


s = ut + ½ at2
v= u + at
v2 = u2 + 2as

Here u is initial velocity & v is final velocity

The Attempt at a Solution



The information of problem as follows
....The hoist......pet rock
u ....2.03....... ?
v...2.03....... ?
s ...2.5 + x......2.5 + x
t ...?....... ?
a...non.......-10

the both time & displacement should be equal

So the displacement of hoist
2.5 + x

The get minimum initial velocity the final velocity should be 0.
v = u + at
0 = u - 10t
t = u/10

the u is equal to

v2 = u2 + 2as
u2 = 20(2.5+x)

the displacement is equal to

s = ut + ½ at2

2.5 + x = ut - 5t2

from here i don't have any idea, please help me.

What is the time of hoist when pet rock reach

2.5 + x = 2.03t
t = (2.5 + x)/2.03

or

x = 2.03t
t = x/2.03

which one i should take
 
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Well I would try to use conservation of energy.

When the naturalist is at the height 2.50m she is traveling at 2.03m/s, so she has a total energy associated with that. Now in order for the rock to reach that high, it must be imparted with kinetic energy equal to her total energy at that height.
 


If you're looking for the absolute minimum initial velocity, then at the end its velocity will be zero because at the apex of its flight the rock will be stationary for a split second where the naturalist can grab it.
 


rock.freak667 said:
Well I would try to use conservation of energy.

When the naturalist is at the height 2.50m she is traveling at 2.03m/s, so she has a total energy associated with that. Now in order for the rock to reach that high, it must be imparted with kinetic energy equal to her total energy at that height.

WOW!
I had previously approached the prob a bit like kalupahana(and got the answer) did but this is much better!
Thanks!
 


The legend said:
WOW!
I had previously approached the prob a bit like kalupahana(and got the answer) did but this is much better!
Thanks!

Normally, for these kinds of problems, I try to use conservation of energy unless I am directly asked to find a time parameter.
 

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