A wee bit of physics.....?

Question

A mountain climber stands at the top of a 50.0 m cliff hanging over a calm pool of water. The climber throws two stones vertically 1.4 s apart and observes that they cause a single splash when they hit the water. The first stone has an initial velocity of 2 m/s. (Assume the positive direction is upward.)

(a) How long after release of the first stone will the two stones hit the water?

(b) What is the initial velocity of the second stone when it is thrown?

(c) What will the velocity of each stone be at the instant both stones hit the water?
first stone

d) second stone


PS....If youre going to tell me to go get help and just do the work yourself or sucking it up and not doing it right......go outside, find a road or trail and go take a hike. Ive been working hard on all these problems trying out various equations so dont tell me I should do it myself because I have!

Neil- The initial velocity you got was not correct. Im not sure where but some place u there is an error. I know this because my homework is done online and I can check answers.

The answer to A was 3.4....

Answer 1

lets try to understand the problem..
stone 1:
initial velocity = 2m/s
time of throw = t (assume)
distance travelled = 50 m

stone 2:
initial velocity = x (to be found out)
time of throw = t + 1.4
distance travelled = 50 m

Now since they both fall into the pool at the same time, lets call that t1
Distance travelled by 1st stone in the time t1-t
=initial velocity * time + gravitational acceleration * time sqared /2
= 1.4(t1-t) + 10[(t1-t)^2]/2
= 1.4(t1-t) + 5(t1-t)^2 = 50

Distance travelled by 2nd stone in the time t1-t-1.4
=initial velocity * time + gravitational acceleration * time sqared /2
= x*(t1-t-1.4) + 10[(t1-t-1.4)^2]/2
= x*(t1-t-1.4) + 5(t1-t-1.4)^2 = 50

Now assume t1-t=y
So you have two equations
1) 1.4y + 5y^2 = 50
2) x(y-1.4) + 5(y-1.4)^2 = 50

Solve the 1st quadratic equation and take its +ve root, you get
y=3.025
Replacing it in the 2nd equation, we get x = 22.640

So I solved (a) and (b) for you
Hopefully, (c) and (d) should be now easy

Answer 2

You need to consider the right equation for the job. Since you're given the initial velocity of the first stone, the initial displacement of the first stone (height of cliff), end displacement of the stone (the water), the acceleration, and you have to find out the time, what equation relates all these and only these variables? From this you can find out the time. Then you can use the answer to the first part to answer the second, then the third and fourth.

Answer 3

doesnt actually work...i plugged and chugged and the answers dont match....just thought i throw that out there