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
1.find the time of flight
2.what is the initial velocity
3.find the angle of the projectile
1 answer:
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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 remains the same in time. The distance x is computed as
.
(a)
The vertical component of the velocity starts from zero and gradually starts to increase due to the acceleration of gravity as follows
This means the vertical height is computed by
Where 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
(b)
We now compute the horizontal distance knowing