A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained
1 answer:
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At a distance of 469.2 m from the original point below the airplane.
Explanation:
First of all, we have to calculate the time it takes for the sack to reach the ground.
To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:
where, taking downward as positive direction:
s = 300 m is the vertical displacement
u = 0 is the initial vertical velocity
t is the time
is the acceleration of gravity
Solving for t, we find the it takes for the sack to reach the ground:
Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:
We also know the total time of flight,
t = 7.82 s
Therefore, we can find the horizontal distance travelled by the sack:
So, the sack will land 469.2 m from the original point below the airplane.
Learn more about free fall and projectile motion:
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