How Fast Was the Second Stone Thrown to Match the First at the Cliff Base?

  • Thread starter Thread starter Ab17
  • Start date Start date
  • Tags Tags
    Fall Free fall
AI Thread Summary
A stone is dropped from a 200m cliff, while a second stone is thrown upwards 1.5 seconds later, both hitting the ground simultaneously. The first stone's velocity after 1.5 seconds is calculated to be -14.7 m/s, and its fall time is determined to be approximately 4.89 seconds. The second stone's upward velocity is derived from the equation of motion, leading to a calculation of -16.94 m/s, which raises concerns about the sign convention. The discussion emphasizes the need to relate the time of flight of the second stone to its initial velocity, considering the upward motion and subsequent fall. Accurate calculations and understanding of motion principles are crucial for solving the problem effectively.
Ab17
Messages
99
Reaction score
2

Homework Statement



a stone is dropped from a cliff 200m high.if a second stone is thrown verticall upwards 1.50seconds after the first was released.strike the cliff base at the same instant as the first stone.with what
velocity was the second stone thrown

2. Homework Equations

Xf=xi +vt + 0.5t^2

Vf= vi + at

The Attempt at a Solution


The First Stone.

d1 = -0.5g*t^2 = -4.9*(1.5)^2 = -11 m. =

Distance traveled by 1st stone after 1,5s. Vf= Vi+at.

Vf= 0 -9.8*1.5 = -14.7m/s

V = -14.7 m/s = Velocity of 1st stone after 1.5 s.
d1 = Vo*t + 0.5g*t^2

-200 = -11 - 14.7 t -4.9t^2

-4.9t^2 -14.7t +189=0

Using Quadratic Formula.

Tf = 4.89 s. = Fall time or time for each stone to reach Gnd. The 2nd Stone.

d = Vo*t + 0.5g*t^2 = -200 m.

Vo*4.89 -4.9*(4.89)^2 = -200

Vo= -16.94m/sWhy Am I getting a negative answer the upward velocity should be positive (the top of the cliff is taking as the X=0 position)
 
Physics news on Phys.org
Ab17 said:
d1 = Vo*t + 0.5g*t^2

-200 = -11 - 14.7 t -4.9t^2

-4.9t^2 -14.7t +189=0

Using Quadratic Formula.

Tf = 4.89 s. = Fall time or time for each stone to reach Gnd.The 2nd Stone.

d = Vo*t + 0.5g*t^2 = -200 m.

Vo*4.89 -4.9*(4.89)^2 = -200

Vo= -16.94m/sWhy Am I getting a negative answer the upward velocity should be positive (the top of the cliff is taking as the X=0 position)

i think you should check your calculation ...
if the time taken by the 2nd stone to go up and come down is known then it should be equal to 2.V0/g as a body being sent up will come to rest after a time V0/g and fall on ground taking same time.
taking
 
Ab17 said:
a stone is dropped from a cliff 200m high.if a second stone is thrown verticall upwards 1.50seconds after the first was released.strike the cliff base at the same instant as the first stone.with what
velocity was the second stone thrown
try to think on the event
1. the 1st. stone is falling through height h so one can calculate time taken say t1 easily using distance traversed.
2. after 1st. one is dropped the second has been thrown up say after 1.5 sec.
3. time of flight of second stone can be easily computed as function of initial velocity of throw.
the above time of flight can be related.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top