Question

If a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower?

Answer 1

Assuming "dropped" means "released" (i.e. initial velocity 0) we can find the height from the following equation:
\[h = 0.5*g*t^2\]
plugging in the values we get: \[h = 0.5*10*3.5^2 = 61.25 (m)\]

Answer 2

answer choices are
a)56.00 meters
b)60.03 meters
c)62.08 meters
d)62.50 meters

Answer 3

plugging in g = 9.8 instead of 10:
\[h = 0.5*9.8*3.5^2 = 60.025 \approx 60.03\]
that's b.
I probably had an arithmetic mistake up there.

Answer 4

nope, simply the difference between g = 10 and 9.8...

Answer 5

A feather is dropped onto the surface of the moon. How far will the feather have fallen if it reaches the surface in 9.00 seconds?
Given: g on moon = -1.63 meters/second2

A)14.0 meters
B)14.7 meters
C)16.00 meters
D)66.0 meters

Answer 6

Again, assuming it was dropped from rest and by the formula:
\[y = y_0 + v_0*t + 0.5*g*t^2\]
with:
\[y=? ; y_0 = 0 ; v_0 = 0; g = 1.63(m/\sec^2)\]
(I took plus instead of minus since I'm taking the positive direction as being down)
we get:
\[y = 0.5*1.63*9^2 = 66.015 (m)\]
That's D.