Question

A student stands at the edge of a cliff and
throws a stone horizontally over the edge with
a speed of 18.9 m/s. The cliff is 77.2 m above
a flat, horizontal beach. At impact, what is its speed?

I calculated the time it takes to reach the bottom, which was 3.967~ seconds. I figured the speed would just be 18.9 m/s, the same as the original speed, but something tells me that's not right. Can someone point me in the right direction please?

Answer 1

Here's a diagram

Answer 2

the horizontal speed would be 18.9m/s .. the vertical speed initially is 0.. due to gravity there will be some final vertical velocity. find that and then vector sum it..

Answer 3

I'm not sure how to vector sum, this is my 2nd week of physics. But I think I would take the 18.9 and just add it to the gravity, which is usually 9.81, so would it just be the sum: 28.71, or would i subtract them? I honestly am not sure...

Answer 4

umm what you gotta do is find the final vertical velocity using v^2-u^2=2as.. here u is the initial velocity which is equal to zero.. so v^2=2as=2*10*77.2.. let this be equal to X^2.. and let the final horizontal velocity be equal to Y. you know this is equal to 18.9.. find Y^2

now the final vertical velicty and the final horizontal velocity are perpendicular to each other.. to find a resultant.. you do this.. sqrt(X^2 + Y^2).. this gives you the answer..

Answer 5

Alright, so bear with me, the equation should be
18.9 = sqrt(x^2+y^2)
18.9 = sqrt(39.29^2+y^2)
357.21 = 1543.999968 + y^2
-1186.799968 = y^2
-34.449963 = y, which is the speed in m/s?

Answer 6

no no no.. you have to find sqrt(x^2 + y^2).. you already have the value of y^2.. which is equal to v^2.. the final vertical velocity..
so the answer would be... sqrt [ 18.9^2 + (2*10*77.2) ]

Answer 7

Alright, I solved for that and got 43.60286688. But it says it's wrong.

Answer 8

43.247.. is that the answer?

Answer 9

Ah! It was right, it was just an error in the online program. I cannot thank you enough for your help:)

Answer 10

Cheers. :)