Calculating Initial Velocity in Vertical Projectile Motion

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SUMMARY

The discussion focuses on calculating the initial velocity of a second stone thrown vertically downward from a 50m cliff, given that the first stone was thrown with an initial speed of 2 m/s and both stones hit the water simultaneously. The first stone takes approximately 0.37 seconds to reach the water, while the second stone, thrown 1 second later, must have an initial velocity of 15.2 m/s downward to reach the water at the same time. The calculations utilize kinematic equations, specifically Vf² = Vi² + aΔy and Vf = Vi + at, with acceleration due to gravity set at 9.8 m/s².

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of vertical projectile motion concepts
  • Familiarity with acceleration due to gravity (9.8 m/s²)
  • Ability to perform basic algebraic manipulations
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn about the effects of air resistance on projectile motion
  • Explore advanced topics in vertical motion, such as maximum height and time of flight
  • Practice problems involving multiple objects in motion with different initial velocities
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This discussion is beneficial for physics students, educators, and anyone interested in mastering concepts of vertical projectile motion and kinematics.

GrandMaster87
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Homework Statement


A mountain climber stands at the top of a 50m cliff that overhangs a calm pool. She throws two stones vertically download 1s apart and observe that the two stones reach the water simultaneously after a while. The first stone was thrown at an initial speed of 2m[tex]^{-1}[/tex].s

Calculate the initial speed at which she threw the second stone. Ignore the effects of friction


Homework Equations





The Attempt at a Solution


Vf[tex]^{2}[/tex]=V[tex]_{i}[/tex][tex]^{2}[/tex] + a[tex]\Delta[/tex]y
=2(2)[tex]^{2}[/tex]=2(9.8)(50)=31.37
Vf=5.6m.s[tex]^{-1}[/tex]

V[tex]_{f}[/tex]=v[tex]_{i}[/tex]+at
5.6=2+9.8t
t=0.367...
t=0.37s

I am not sure if I am right...because now i have to -1 second from the t above hence giving me a negative second to enter in a formula to find the initial velocity of the stone...

Can someone help me...the model answer gives me an anser of 15.2m/s downwards...how can that be? surely it is wrong
 
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Find how long it takes the first rock to reach the water, that also will let you easily find the time the second rock was in the air. Then just solve for Vo.
 
i did an i get 0.37s...and if the the second stone was released 1 s later then how can that be?
 

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