A stone is dropped into a river from a bridge 43.9m above the water.?

Question

Another stone is thrown vertically down 1.00s after the first is dropped. The stones strike the water at the same time.
a) what si the initial speed of the second stone?
b) plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released. You can just e-mail this to me .

Please don't just give answers but show me how becuse I really want to learn how to do it.

Thanks

Answer 1

This is not too bad. You really only need two equations and those are:
s = v0*t + (1/2)*g*t²
v(t) = v0+gt
where s is distance traveled, v0 is initial velocity, v(t) is velocity at time t, g is acceleration of gravity (9.8 m/s²) and t is time in seconds.

When the first stone is released, it's initial velocity is 0 and it begins to accelerate at 9.8 m/s². So
43.9 = (1/2)*9.8*t² and, solving for t, t = 2.993 sec. until the stone hits the water.

The 2'nd stone hits the water at the same time (and travels the same distance) but it does so in 1.993 seconds. So
43.9 = v0*1.993 + (1/2)*9.8*1.993² and solving for v0 we get v0 = 12.261 m/s

How to plot velocity vs time: The fastest stone is travelling at 12.261 + 9.8*1.993 = 31.792 m/s. Most graph paper has at least 30 squares along the short dimension, so call each square 1 m/s and, along the long dimension, maybe every 10 squares = 1 second.

Now just use the 2'nd equation to get velocities for each stone maybe every half second. Plot the points on the graph paper and you're all done.


Doug

Answer 2

In order for the second stone to catch up to the first and land at the same time, it must make up the 1 second delay over the distance of the fall, which is the height of the bridge. Both stones will have the same acceleration and time to reach bottom due to gravity, The initial velocity of the second stone will just add to that. So the initlal velocity must be fall distance/time delay.

For the plot of velocity vs time, the velocity due to gravity acceleration is linearly increasing (v=g*t). The first stone's linear plot will start at 0, but the second won't have any velocity until t=1sec, and then it will have a velocity of the initial velocity you computed above. From then on the velocity will be a linear increase.

I hope this gives you enough information to solve the problem,

Answer 3

About the first stone: Time to hit the water can be found using
y = Vo*t + 0.5*a*t^2
where y is the distance to the water, Vo = original velocity = 0, a (acceleration due to gravity) = 9.8 m/s^2, and t is your unknown, time.

About the second stone: Time to hit the water is t (from above) less 1 second. You can use the same equation as above:
y = Vo*t + 0.5*a*t^2
where y is the same, Vo is now your unknown, t is your above result -1 s, and a is still 9.8 m/s^2.

I'm sure you can do the math. Your plots will be straight lines. All you need is to know 2 points on the graph for each rock and draw a line between them. You know the initial velocity of each rock, that's the starting points for your lines, and the times to the water. Plug those into the equation
V = Vo +a*t,
once for each rock and you have the beginning and end points for your lines.

Answer 4

an undemanding physics question. S = Vo*t + a million/2 a*t^2 study distance = preliminary speed situations Time plus a million/2 acceleration situations Time squared. If there have been no acceleration, then speed * time = 40 could be perfect. however the rock became thrown down and is being acted on by ability of gravity (32.2 ft/sec^2 or 9.8 m/sec^2) Equation will become S = (10)*4 + (a million/2)*9.8*(4^2) = 118.4 meters.

Answer 5

You start out using the equation:

y = y0 + (v0 * t ) - (½ * g * t²)

You've probably seen it written with x's instead of y's. Anyway, y0 is the starting height (43.9 m), y is the ending height (0 m), v0 is the initial velocity, t is the time, and g is gravitational acceleration. First you should put in the values you know for the stone that is dropped, and solve for t.

0 = 43.9 + 0 - (½ * 9.81 * t²)
t² = (43.9 * 2) / 9.81
t = 2.9917 s

You know that the travel time for stone 2 is one second shorter than the travel time for stone 1. So t2 = (t1 - 1) = 1.9917 s.

y = y0 + (v0 * t ) - (½ * g * t²)
0 = 43.9 + (v0 * 1.9917) - (½ * 9.81 * 1.9917²)

v0 = -12.2722 m/s,
or
v0 = 12.2722 m/s downward.

Answer 6

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