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)?
+130466 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.
+118703 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
¨y=−9.8m/s˙y=−9.8ty=−9.8t22+61.6
Find t when y=0
−61.6×2=−9.8t2t≈3.545621042
˙y=−9.8×3.545621042=−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. 
+130466 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.
