A stone is thrown horizontally from the top of a 25.00m cliff the stone lands at a distance of 40.00 m from the edge of the cliff What is the initial horizontal velocity?

Answer 1

The short answer is 28 m/s.

Here is how you come to this. There is a basic physics formula you use to determine the distance something has traveled, using your initial distance/position, initial velocity, acceleration and time. This formula is:

x = x0 + v0t + 1/2at^2

(x0 is x-naught, the initial distance, v0 is v-naught, the initial velocity. The zero should be a subscript).

The formula for it's velocity is simply the first derivative of the distance formula, which is:

v = v0 + at

To find the velocity it is at after it has traveled 40m, we first need to know the time it took to fall 40m, so we can make use of our distance formula. We know that we have traveled 40m, so we use that for x. We can say our initial position is 0m and our initial velocity is 0m/s. Our acceleration is that of gravity, which is 9.8m/s/s. We plug these in and solve for t:

40 = 0 + 0t + 1/2(9.8)t^2

40 = 4.9t^2

t = sqrt(40/4.9) = approximately 2.86

Then, we just use that time and the same acceleration and plug them into our velocity formula:

v = 0 + 9.8(2.86)

v = 9.8(2.86)

v = 28 m/s

(It's exactly 28 m/s if you use the exact value for t)

Answer 2

25m = 1/2gt^2 so 25 = 1/2gt^2 g = 9.8 m/s^2 1/2g = 4.9 25/4.9 = t^2 = 5.1, sq-root 5.1 = t = 2.2587... secs 40m /2.2587 = horizontal velocity = 17.7 m/s