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fenix001 [56]
3 years ago
13

An object was thrown at a certain angle above the ground.It reaches a maximum height of 42.50 meters and hits back the ground 76

meters
1.find the time of flight
2.what is the initial velocity
3.find the angle of the projectile

Physics
1 answer:
aksik [14]3 years ago
7 0

Answer:

Explanation:  Please see my attached calculations.

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A fighter bomber is making a bombing run flying horizontally at 500 knots (256m/s) at an altitude of 100.0m.
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Answer:

<em>(a) t = 4.52 sec</em>

<em>(b) X = 1,156.49 m</em>

Explanation:

<u>Horizontal Launching </u>

If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity v_o remains the same in time. The distance x is computed as .

\displaystyle x=v_o.t

(a)

The vertical component of the velocity v_y starts from zero and gradually starts to increase due to the acceleration of gravity as follows

v_y=gt

This means the vertical height is computed by

\displaystyle h=h_o-\frac{gt^2}{2}

Where h_o is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0

\displaystyle t=\sqrt{\frac{2h_o}{g}}

\displaystyle t=\sqrt{\frac{2(100)}{9.8}}=4.52\ sec

(b)

We now compute the horizontal distance knowing v_o=256\ m/s

\displaystyle x=(256).(4.52)

X=1,156.49\ m

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