Posted by ruby on Monday, July 21, 2008 at 8:52pm.
Suppose that a person is standing on the edge of a cliff 50 m high, so that the ball in his/her hand can fall to the base of the cliff.
Questions
(a)How long does it take to reach the base of the cliff?
(b) What is the total distance traveled by the ball?
Attempted Answers
(a) I got 40.2 s because I divided the total distance the ball covered (50 m) by the pull of gravity on the object, 9.8 m/s. However I feel like my answer is wrong, can someone please correct this for me.
(b) The total distance traveled by the ball would remain to be 50 m because the ball was dropped down a cliff 50 m high, and there was no going up or down involved. I think this one is right
*** Also, i attempted to solve these questions mentally, however i don't know how to solve them step by step using equations, so can some one please demonnstrate this for me? Thanks for all the help. =)

physics  GK, Monday, July 21, 2008 at 9:13pm
The equation is:
50 m = (1/2)(9.8m/s^2)(t^2)
(It is less confusing to use absolute values hereNo negative signs)
t = sqrt[(2*50m)/(9.8m/s^2)] = sqrt(10.2s^2)
(t = much shorter than 40.2 s)

question?  ruby, Monday, July 21, 2008 at 9:45pm
so, then to calculate the distance, the formula would be 50 m =(1/2)(9.8m/s^2)(t^2) and you would neglect the negative signs, but then instead of dividing 50m by 90.8m/s^2, how would u use that formula to find the value of t? Sorry, i am just really confused regarding all of this. =( you see for these questions we were to use this equation
Y = Yo + Vot + 1/2gt^2
But how would you plug that in? Thank you for all the help though, i really appreciate it. =)

physics  Damon, Monday, July 21, 2008 at 9:50pm
Yo = 50
Vo = 0 (does not throw it up or down, just drops it)
(1/2) g = 4.9
Y = 0 at the bottom
so
0 = 50  4.9 t^2
or
4.9 t^2 = 50
t^2 = 10.2
t = 3.2 seconds

physics  GK, Monday, July 21, 2008 at 10:45pm
Yours (Damon's) is a more elegant solution than mine, since you are using a more general formula for vertical motion and correct algebraic signs. I assume the relationship used is:
Y = Yo + (1/2)gt^2, using the values you assigned. The zero point is the bottom of the cliff.
An equivalent formula would be:
Y = (1/2)gt^2, with:
Y = 50m, g = 9.8m/s^2
Since the displacement is traversed in the direction of motion, Y is negative. This time the zero point is the starting point.
How we choose a reference point makes a difference but should not affect the final answer if the algebraic signs are consistent with our choice.

physics  ruby, Tuesday, July 22, 2008 at 6:12am
thanx =) for all your help!
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