Projected motion/trajectory question [EASY]

[ujellytek]

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

 

[gneill]

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.

 

[ujellytek]

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

 

[gneill]

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.

 

[ujellytek]

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

 

[gneill]

That certainly looks better!