If there were no air resistance, a quarter dropped from the top of New York's empire state building would reach the ground 9.6 seconds later, what would its speed be? How high is the building?
distance = 1/2*g*t^2
velocity = g*t
I don't know how to use those formulas to figure out what the speed and how high the building would be ?
distance = 1/2*g*t^2
distance is what you want.
1/2 is obvious.
g = 9.8 meters/secnd/second
t = time = 9.6 seconds in the problem. Solve for d.
Okay, thank you! I'm pretty sure I got it! What about to know how high the building is though? Sorry if you already answered that too, I am awful with this kind of stuff
The distance IS how high.
To find the speed of the quarter when it reaches the ground without air resistance, we can use the equations of motion for free fall:
The equation to find speed is given by:
v = u + gt
Where:
v = final velocity (speed)
u = initial velocity (0 m/s in this case, as the quarter is dropped)
g = acceleration due to gravity (9.8 m/s^2 on Earth)
t = time (9.6 seconds in this case)
Substituting the given values into the equation, we have:
v = 0 + (9.8 m/s^2)(9.6 s)
v = 94.08 m/s
Therefore, the speed of the quarter when it reaches the ground without air resistance would be approximately 94.08 meters per second.
Now, to calculate the height of the Empire State Building, we can use another equation of motion:
h = (1/2)gt^2
Where:
h = height
g = acceleration due to gravity (9.8 m/s^2 on Earth)
t = time (9.6 seconds in this case)
Substituting the given values into the equation, we have:
h = (1/2)(9.8 m/s^2)(9.6 s)^2
h = 446.256 meters
Therefore, the height of the Empire State Building is approximately 446.256 meters.