An inquisitive physics student and mountain climber climbs a 50.0m high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously? (c) What is the speed of each stone at the instant the two stones hit the water?
(a) X=-2t-4.9t^2
50=-2t-4.9t^2
t=2.99s
(b) X=V(t-1)-4.9(t-1)^2...
The thing that I don't understand is that the question states that both stones hit the ground at the same time (splash together), but in solving this problem we use "t-1" which would mean that the second stone hits the ground one second before the first stone. Am I interpreting the question wrong. Could you please explain fully
ok, then you know the second stone hit the water in 1.99sec after being thrown.
50=V(1.99)-4.98(1.99)^2
solve for V
Now the speed of each:
first: vf=vi+gt solve that , t=2.99
second: vf=V+gt
t=1.99
Certainly! Let's break down the problem and understand it step by step.
We are given that the first stone is thrown vertically downward with an initial speed of 2.00 m/s and that the two stones cause a single splash in the calm pool of water below. We need to find the answers to the following questions:
(a) How long after the release of the first stone do the two stones hit the water?
(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?
(c) What is the speed of each stone at the instant the two stones hit the water?
To solve part (a), we need to find the time it takes for the first stone to hit the water. We can use the equation of motion for vertically downward motion, which is given by:
x = ut + (1/2)at^2
Here, x is the vertical displacement (in this case, the height of the cliff) which is 50.0 m, u is the initial speed (2.00 m/s), t is the time, and a is the acceleration due to gravity (-9.8 m/s^2). Plugging in the values, we get:
50 = -2t - 4.9t^2
Simplifying and rearranging the equation, we find a quadratic equation:
4.9t^2 + 2t - 50 = 0
Solving this equation, we find t ≈ 2.99 s. This is the time it takes for the first stone to hit the water after its release.
Now let's move on to part (b). We need to find the initial velocity of the second stone such that both stones hit the water simultaneously. Since the second stone is released 1.00 s after the first stone, we need to consider its motion with respect to this time difference.
We can use the same equation of motion as before, but this time we need to account for the time difference of 1.00 s. The equation becomes:
x = V(t-1) - 4.9(t-1)^2
Here, x is still 50.0 m, V is the initial velocity of the second stone, and t-1 is the time difference considering the second stone is released 1.00 s after the first stone. Plugging in the values, we get:
50 = V(t-1) - 4.9(t-1)^2
Simplifying and rearranging, we again end up with a quadratic equation, and this time it involves the initial velocity V of the second stone. Solving this equation, we find the value of V that ensures both stones hit the water simultaneously.
Finally, for part (c), once we have the time at which both stones hit the water (from part a), we can substitute this time value into either of the equations of motion to find the speeds of the stones at that instant.
I hope this explanation clarifies the problem for you. Let me know if you have any further questions!