Question

A falling stone is at a certain instant 128 feet above the ground. 2 seconds later it is only 16 feet above the ground. From what height was it dropped?

Answer 1

can someone please help me?

Answer 2

us this Newtonian equation:
\[\Delta y = V_0 t + \frac{ 1 }{ 2 }at^2\]

Answer 3

a = gravity = -32 ft/s^2

Answer 4

solving it on paper atm

Answer 5

ok

Answer 6

the answer choices are 139 138 137 140 and 135

Answer 7

\[\Delta y = y_2 - y_1\]
since the stone is dropped, the initial velocity = 0.

here are the equations:
\[128 - y_0 = 0 - 16t^2\]\[16-y_0 = 0 - 16(t+2)^2\]
equate them both for y_0 and solve for t. then you an solve for y_0 by plugging t in.

really sorry i gotta go. i hope you can solve from this ^_^ o

Answer 8

but waht is y0?

Answer 9

That is the initial height, what you are trying to solve. Do as @Euler271 posted and you will come up with one of the choices..........Its there just do it.