This is two different questions.. answer both please.?

Question

1)How far will a freely falling object have fallen from a position of rest when its instantaneous speed is 24 m/s?

2)For a freely falling object dropped from rest, what is the acceleration at the following times?

a)the end of the fifth second?
b)the end of the sixth second?
c)the end of the any elapsed time?

Answer 1

For the first question, under constant accelleration, v=at.

Here, 24m/s = 9.8m/s^2 x time.

Time = 24/9.8

Distance = 1/2 at^2 = 1/2(9.8)((24/9.8)^2) = 1/2(24)(24/9.8)
= 12 (24/9.8) = ??? You do the math.

Well, the second question is not a trick question, exactly, but it designed to see if you get it.

Do you???

What is accelleration???

Is it the CONSTANT g - usually approximated with 9.8 m/s^2???

What is the constant after five seconds???
What is the constant after six seconds???

Answer 2

1) V(initial) = 0 m/s
V(final) = -24 m/s (falling means negative)
a = -9.8 m/s^2 (acceleration due to gravity)
d = unknown

v(final)^2 - v(initial)^2 = 2 * a * d
(-24)^2 = 2 * -9.8 * d
d = -29.4 m

2) the only acceleration acting on a free falling object is due to gravity. Acceleration due to gravity on Earth is -9.8 m/s^2