An inquisitive physics student and mountain climber climbs a 50.0 m 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 2.08 m/s downward.
(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 they are to hit simultaneously?
2 m/s (downward)
(c) What is the speed of each stone at the instant the two hit the water?
--first stone? (downward)
--the second stone? (downward)
I'd say they would never hit the water at the same time because the first stone had a initial DOWNWARD velocity of 2.08 m/s. The second stone lost one second plus it started at zero vertical velocity.
If the situation is reversed, if the second stone is given an initial downward velocity of 2.08 m/s, there is a chance they may hit the water together.
(1)
Using
S=vertical distance travelled, m
t = time in seconds since release of first stone.
S = 0*t+(1/2)(-g)t² = 2.08*(t-1)+(1/2)(-g)(t-1)²
Solve for t
(2)
v(first) = 0+(-g)*t
v(second) = 2.08+(-g)*(t-1)
To answer these questions, we need to use the equations of motion and apply them to the problem scenario.
(a) How long after release of the first stone do the two stones hit the water?
Let's assume the time it takes for the first stone to hit the water is t seconds. Since the first stone is in free fall, we can use the equation:
y = y0 + v0t + (1/2)at^2
In this case, y0 represents the initial height (50.0 m), v0 represents the initial velocity (2.08 m/s downwards), and a represents the acceleration due to gravity (-9.8 m/s^2). As both stones hit the water, their final height will be zero. Therefore, we can write the equation for the first stone as:
0 = 50.0 + (2.08)t + (1/2)(-9.8)t^2
Simplifying this equation, we get:
-4.9t^2 + 2.08t + 50.0 = 0
We can solve this quadratic equation to find the value of t. Using the quadratic formula, t is approximately 4.019 seconds.
(b) What initial velocity must the second stone have if they are to hit simultaneously?
Since the two stones are released 1.00 second apart and they hit the water at the same time, the time of flight for the second stone should be 4.019 - 1.00 = 3.019 seconds. Let's assume the initial velocity of the second stone is v2 m/s downwards.
Using the same equation of motion as before, for the second stone, we have:
0 = 50.0 + v2(3.019) + (1/2)(-9.8)(3.019)^2
Simplifying this equation, we can solve for v2. The approximate value of v2 is -23.68 m/s, which means the second stone should have an initial velocity of approximately 23.68 m/s downwards.
(c) What is the speed of each stone at the instant the two hit the water?
To find the speed, we can use the equation:
v = v0 + at
For the first stone, v0 is given as 2.08 m/s downwards, and the acceleration (due to gravity) is -9.8 m/s^2. Plugging these values into the equation, we find that the speed of the first stone at the instant it hits the water is approximately 38.84 m/s downwards.
For the second stone, v0 is -23.68 m/s downwards, and the acceleration (due to gravity) is still -9.8 m/s^2. Using the same equation, we find that the speed of the second stone at the instant it hits the water is approximately 12.93 m/s downwards.