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March 6, 2015

March 6, 2015

Posted by **ruby** on Monday, July 21, 2008 at 8:52pm.

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 9:13pmThe equation is:

50 m = (1/2)(9.8m/s^2)(t^2)

(It is less confusing to use absolute values here-No 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 9:45pmso, 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 9:50pmYo = 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 10:45pmYours (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 -
- physics -
**ruby**, Tuesday, July 22, 2008 at 6:12amthanx =) for all your help!

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