On Mars, the acceleration due to gravity is 3.71 m/s2. How far would a rock fall in 3 s if you dropped it on Mars?
Knowns:
acceleration (a): -3.71 m/s^2
time (t): 3 sec
initial velocity (Vi): 0 m/s
*Since you dropped it, there is no force put into making the ball fall. That's why the initial velocity is 0.
Unknown:
Fall distance(Δy):?
Δy=Vit+(1/2)at^2
Δy=0+(1/2)(-3.71)(3)^2
Δy=-16.695m
The ball would fall 16.695m
Well, on Mars, gravity sure likes to take its time! With an acceleration due to gravity of 3.71 m/s², if you dropped a rock on the red planet, it would only fall for 3 seconds.
Let me crunch the numbers and whip out my Martian calculator. *beep boop beep*
Using the formula for distance fallen (d) = 1/2 * a * t², where a is the acceleration due to gravity and t is the time, we can calculate the distance.
So, plugging in the values, we get:
d = 1/2 * 3.71 m/s² * (3 s)²
d = 1/2 * 3.71 m/s² * 9 s²
*dusts off calculator*
d = 1/2 * 3.71 m/s² * 81 s²
*dramatic drumroll*
d = 1/2 * 298.51 m
*drops calculator*
Voilà! The rock would fall approximately 149.26 meters on Mars in 3 seconds.
To find the distance a rock would fall on Mars in 3 seconds, we can use the formula:
d = (1/2) * g * t^2
Where:
d = distance
g = acceleration due to gravity
t = time
Given that the acceleration due to gravity on Mars is 3.71 m/s^2 and the time is 3 seconds, we can substitute these values into the formula:
d = (1/2) * 3.71 m/s^2 * (3 s)^2
First, we square the time:
d = (1/2) * 3.71 m/s^2 * 9 s^2
Next, we multiply the acceleration due to gravity by 9:
d = (1/2) * 3.71 m/s^2 * 9
Now, we can simplify the equation further:
d = 16.695 m
Therefore, a rock would fall approximately 16.695 meters if dropped on Mars for 3 seconds.
To determine how far the rock would fall on Mars in 3 seconds, we can use the equation for displacement:
\[d = \frac{1}{2} \times g \times t^2\]
Where:
d is the displacement (distance)
g is the acceleration due to gravity on Mars (3.71 m/s^2)
t is the time (3 seconds)
Substituting the given values into the equation, we have:
\[d = \frac{1}{2} \times 3.71 \times (3^2)\]
Now let's calculate the result:
\[d = \frac{1}{2} \times 3.71 \times 9\]
\[d = 1.855 \times 9\]
\[d = 16.695\]
Therefore, the rock would fall approximately 16.7 meters on Mars in 3 seconds.