a penny dropped from the Empire state building. how far does the penny drop in 6 seconds? how fast will it be moving after 10 seconds?
Drag out your good old equations.
s = 1/2 at^2
v = at
Ring a bell?
In feet,
s = 16t^2 = 16*36 = 576ft
v = 32*10 = 320 ft/s
You can use the formulas
y = (g/2) t^2
for the distance it drops and
V = g*t
for the velocity.
t is the time it has fallen.
They are good for the inital several seconds, but after that, air friction slows it down.
g is the acceleration of gravIty, which you should know the vaue of.
To find how far the penny drops in 6 seconds, we can use the equation of motion under constant acceleration:
s = ut + 0.5 * a * t^2
Where:
s is the distance traveled
u is the initial velocity (which is 0 since the penny was dropped)
a is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time (6 seconds)
Plugging in the values into the equation, we get:
s = 0 * 6 + 0.5 * 9.8 * 6^2
s = 0 + 0.5 * 9.8 * 36
s = 0 + 0.5 * 352.8
s = 0 + 176.4
s ≈ 176.4 meters
So, the penny would drop approximately 176.4 meters in 6 seconds.
To find how fast the penny will be moving after 10 seconds, we can use another equation of motion:
v = u + a * t
Where:
v is the final velocity (what we want to find)
u is the initial velocity (0 m/s since the penny was dropped)
a is the acceleration due to gravity (-9.8 m/s^2 since it acts in the opposite direction to the initial velocity)
t is the time (10 seconds)
Plugging in the values into the equation, we get:
v = 0 + (-9.8) * 10
v = -98 m/s
Therefore, after 10 seconds, the penny will be moving downward at a speed of 98 m/s. Note that the negative sign indicates the direction (downward).