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The bullet travels a horizontal distance of 276.5 m
The bullet is shot forward with a horizontal velocity
. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>
The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.
The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.
Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

Substitute 0 m/s for
, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

The horizontal distance traveled by the bullet is given by,

Substitute 500 m/s for
and 0.5530s for t.

The bullet travels a distance of 276.5 m.
Answer:
So airplane will be 1324.9453 m apart after 2.9 hour
Explanation:
So if we draw the vectors of a 2d graph we see that the difference in angles is = 83 - 44.3 = 
Distance traveled by first plane = 730×2.9 = 2117 m
And distance traveled by second plane = 590×2.9 = 1711 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.
Using the law of cosine,
representing the distance between the planes, we see that:

d = 1324.9453 m
Answer: 16N
Explanation:
Given that:
mass of box M= 2 kg
Initial speed V1 = 4 m/s
Final speed V2 = 8 m/s
Time taken T= 0.5 s
Average strength of this force F = ?
Now, recall that Force is the rate of change of momentum per unit time
i.e Force = momentum / time
Hence, F = M x (V2 - V1)/T
F = 2kg x (8 m/s - 4 m/s) / 0.5s
F = 2kg x (4 m/s / 0.5s)
F = 2kg x 8 m/s/s)
F = 16N
Thus, the average strength of this
force is 16 newton.