You drop a penny from the top of the Giant Drop which is 61.6 m tall. What is the velocity of the penny when it strikes the ground (ignore air resistance)?
Here's a slightly easier "formula" to use which doesn't involve the need to solve for "time" first.
v2 = u2 + 2 (a) (s)......where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the initial position. Note that "u" is just "0" in this problem, since we're just "dropping" something. So we have...
v2 = 2 (9.8m/s2) (61.6m) =
v = √(2 (9.8m/s2) (61.6m)) = 34.747 m /s...which is exactly what Melody found, too.
Since the motion is "downward," this, of course, is "negative."
Your choice !!!
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Let me append something to this...Melody's method might be of more interest, because it gives a more complete "picture" of the situation.
You drop a penny from the top of the Giant Drop which is 61.6 m tall. What is the velocity of the penny when it strikes the ground (ignore air resistance)?
Accel = 9.8m/s2, original velocity=0, original displacement=61.6
$$\ddot y = -9.8m/s\\\\
\dot y = -9.8t\\\\
y= \frac{-9.8t^2}{2}+61.6\\\\$$
Find t when y=0
$$-61.6\times2=-9.8t^2\\
t\approx 3.545621042$$
$$\dot y = -9.8\times 3.545621042
=-34.747$$
so the coin will hit the ground with a downward velocity of 34.7 m/s
If you don't understand some of the steps let me know. There could easily be some silly errors. ![]()
Here's a slightly easier "formula" to use which doesn't involve the need to solve for "time" first.
v2 = u2 + 2 (a) (s)......where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the initial position. Note that "u" is just "0" in this problem, since we're just "dropping" something. So we have...
v2 = 2 (9.8m/s2) (61.6m) =
v = √(2 (9.8m/s2) (61.6m)) = 34.747 m /s...which is exactly what Melody found, too.
Since the motion is "downward," this, of course, is "negative."
Your choice !!!
![]()
Let me append something to this...Melody's method might be of more interest, because it gives a more complete "picture" of the situation.