Question

-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. =)

Answer 1

The 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)

Answer 2

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. =)

Answer 3

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

Answer 4

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.

Answer 5

thanx =) for all your help!