How do you calculate the velocity of a quarter dropped from a tower after 10 seconds?
It depends -- there's terminal velocity, peak velocity, average velocity, etc....what exactly are you looking for?
im looking for average velocity
The formula for average velocity is:
position2 - position1
-------------------------
time2 - time1
It looks like there is not enough information to figure out the average velocity of the quarter. You would need the height of the tower to fill in all the variables to successfully solve the equation.
can i have your number so we can text
what is the velocity of auarter dropped from a tower after 10 seconds?
we study these in 7th grade ___ ___
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To calculate the velocity of a quarter dropped from a tower after 10 seconds, you need to use the equations of motion. In this scenario, we will assume that air resistance is negligible.
Step 1: Determine the acceleration due to gravity (g)
The acceleration due to gravity on Earth is approximately 9.8 meters per second squared (m/s^2). This value will be used in the calculations.
Step 2: Calculate the distance fallen (d)
The distance traveled by the quarter after 10 seconds can be calculated using the formula:
d = (1/2) * g * t^2
In this case, t represents the time, which is 10 seconds.
Step 3: Calculate the velocity (v)
The velocity of the quarter can be calculated using the formula:
v = g * t
where g is the acceleration due to gravity (9.8 m/s^2) and t is the time (10 seconds).
Plug in the values:
v = (9.8 m/s^2) * 10 s
v = 98 m/s
So, the velocity of the quarter dropped from the tower after 10 seconds is 98 meters per second (m/s).