Question

E 6.15. A jet directed at 30° reaches a maximum height of 3 m at a horizontal distance of 18 m. Determine the issuing velocity of the jet. (16.9 m/s)

          E 6.15. A jet directed at 30° reaches a maximum height of 3 m at a horizontal distance of 18 m.
Determine the issuing velocity of the jet.
(16.9 m/s)
        

Added by Andrea R.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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E 6.15. A jet directed at 30° reaches a maximum height of 3 m at a horizontal distance of 18 m. Determine the issuing velocity of the jet. (16.9 m/s)
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Transcript

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00:01 The question is, particle is projected from the surface of the earth with the speed of 20 meters per second at an angle of 30 with the horizontal.
00:09 We have to find the time range and height.
00:12 So, in the given we have, they have told us that the speed is, so the velocity is initial is u is equal to 20 meters.
00:24 It is short at an angle or it is projected at an angle of 30 degrees.
00:28 So the theta is 30 degrees.
00:30 And we have to find the time we have to find the range and also we have to find the maximum height on this sum is of projectile motion so we'll be using the formulas of projectile motion so as we know the projectile motion if i draw a projectile motion it is like this so any projectile motion will travel like this so if this is the surface then at this place it is short at the the angle of 30 degrees.
01:06 So it will travel like this and then again come back.
01:08 So, you have to find the time, the range and the height, the maximum height it reaches at the middle point.
01:14 So first of all, the time is calculated by the formula 2 u sine theta upon g...
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