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

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SUMMARY

The problem involves two stones: the first stone is dropped from a height of 200 meters, while the second stone is thrown vertically upwards 1.5 seconds later. The calculations show that the first stone reaches a velocity of -14.7 m/s after 1.5 seconds and takes approximately 4.89 seconds to hit the ground. The second stone must be thrown with an initial velocity of approximately 16.94 m/s to ensure both stones strike the cliff base simultaneously. The negative value for the second stone's velocity indicates a misinterpretation of the upward direction, which should be considered positive.

PREREQUISITES
  • Understanding of kinematic equations, specifically Xf=xi + vt + 0.5t^2 and Vf= vi + at.
  • Knowledge of gravitational acceleration, typically -9.8 m/s².
  • Familiarity with the quadratic formula for solving equations.
  • Basic principles of projectile motion and time of flight calculations.
NEXT STEPS
  • Review kinematic equations in physics to solidify understanding of motion under gravity.
  • Learn how to apply the quadratic formula to solve motion-related problems effectively.
  • Study projectile motion concepts, focusing on upward and downward trajectories.
  • Explore the relationship between time of flight and initial velocity in vertical motion scenarios.
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Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for examples of real-world applications of these concepts.

Ab17
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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)
 
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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.
 

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