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Dmitry [639]
3 years ago
14

A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

bullet's flight time? neglect the air resistance.
Physics
1 answer:
Nezavi [6.7K]3 years ago
4 0

We use the equation of motion for vertical component,

s_{y} = u_{y} t+\frac{1}{2} gt^2.

Here, s_{y}   is displacement of bullet, u_{y}  is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

s_{y} =\frac{1}{2}gt^2

Given, s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m

Substituting the values, we get time of flight

2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s

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