A penny is dropped from the top of a building 310 m high. Ignoring air resistance, if the penny falls straight down, with what speed would it hit the ground?
how long does it take to fall?
4.9t^2 = 310
so, its downward speed is
v = 9.8t
Since the object is in free fall, it experiences uniformly accelerated motion (UAM). We can use the formula:
vf^2 - vo^2 = 2gd
where
vf = final velocity
vo = initial velocity (in this case it's zero because freefall)
g = acceleration due to gravity = 9.8 m/s^2
d = distance or height
We are looking for terminal velocity (the velocity of object just before it hits the ground). Substituting the values:
vf^2 - vo^2 = 2gd
vf^2 - 0 = 2 (9.8) (310)
vf^2 = 6076
vf = 77.95 m/s
hope this helps~ `u`
(I apologize if someone has already posted an answer/solution before me. I've got slow internet and I can't seem to post my comment right away.)
To find the speed at which the penny hits the ground, we can use the equations of motion in one dimension. In this case, we will use the formula for final velocity.
The formula for calculating the final velocity of an object in free fall is given by:
v = √(2gh)
Where:
v = final velocity (speed at which the penny hits the ground)
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height of the building (310 m in this case)
Now, let's plug in the values into the equation and calculate the final velocity:
v = √(2 * 9.8 * 310)
v = √(6076)
v ≈ 77.95 m/s
Therefore, the penny would hit the ground with a speed of approximately 77.95 m/s.