Projected motion/trajectory question [EASY]

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The problem involves two rocks, A and B, dropped from the same height, with rock A dropped first and rock B thrown down after 2 seconds. Rock A travels 19.6 meters in the first 2 seconds, leading to a calculated time of approximately 4.05 seconds to hit the river. The initial velocity needed for rock B to hit the river at the same time as rock A was initially miscalculated, but after correcting for the velocity of rock A at the 2-second mark, the correct initial velocity for rock B is determined to be around 27.38 m/s. The discussion emphasizes the importance of accounting for the velocity of rock A when setting up the equations for rock B. Accurate application of kinematic equations is crucial for solving such trajectory problems.
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Homework Statement


The rock 'A' is dropped 100m above a river. After 2 seconds, rock 'B' is thrown down from the same height. What initial velocity must be given to rock 'B' so that both rocks hit the river at the exact same time? [No air resistance]

Homework Equations


What is the answer?

The Attempt at a Solution


http://imgur.com/a/59fkU[/B]

Rock A: v1=0 During a 2s drop it travels 19.6m, -4.9(2)^2+100
So the equation for rock A is d=-4.9(t)^2+80.4
When will rock A hit the river? isolate t and I get ~4.0507 seconds

Rock B: v1=? equation is easy to set up: d=-4.9(t)^2+v1(t)+100 (starts 19.6m behind the dropped rock)
Now we sub in the time rock A hits the river: (4.9(4.0507)^2-100) * (4.0507)^-1=v1
This gives me an answer of ~-4.849m/s
 
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That's relevant equations, not relevant questions. What common kinematic equations might apply to this problem?

Your attempt at solution image is barely readable if at all. Please type in your attempt.
 
gneill said:
That's relevant equations, not relevant questions. What common kinematic equations might apply to this problem?

Your attempt at solution image is barely readable if at all. Please type in your attempt.
Done
 
ujellytek said:
Rock A: v1=0 During a 2s drop it travels 19.6m, -4.9(2)^2+100
So the equation for rock A is d=-4.9(t)^2+80.4
When will rock A hit the river? isolate t and I get ~4.0507 seconds
I think you forgot the velocity that rock A had attained by the time it reached the 2 second mark when you wrote its equation. I think you'll find that the time remaining for rock A to hit the water from the 2 second point is much less than 4 seconds.
 
gneill said:
I think you forgot the velocity that rock A had attained by the time it reached the 2 second mark when you wrote its equation. I think you'll find that the time remaining for rock A to hit the water from the 2 second point is much less than 4 seconds.
You're right, thanks. The velocity then turns out to be ~27.38m/s instead of 4.849m/s
 
That certainly looks better! :smile:
 
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