Answer:
Distance traveled during this acceleration will be 6950 m
Explanation:
Wear have given maximum speed tat will be equal to final speed of the car v = 278 m/sec
Constant acceleration
As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec
From third equation of motion
Putting all values in equation
s = 6950 m
So distance traveled during this acceleration will be 6950 m
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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