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A projectile is launched at a speed Vo = 31 m/s and angle θo = 40°. Please answer the following: A. When is the projectile at a horizontal distance x = 75 m? B. What is its height at that time?

          A projectile is launched at a speed Vo = 31 m/s and angle θo = 40°.
Please answer the following:
A. When is the projectile at a horizontal distance x = 75 m?
B. What is its height at that time?
        

Added by Terri W.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A projectile is launched at a speed Vo = 31 m/s and angle θo = 40°. Please answer the following: A. When is the projectile at a horizontal distance x = 75 m? B. What is its height at that time?
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Transcript

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00:01 Hello, here projectile was fired at the initial speed of 31 meters per second at an angle of 40 degree with respect to the horizontal.
00:09 And first, we have to find time when the projectile horizontal distance is 75 meters.
00:16 Let's gauge this flight.
00:25 So here the projectile travels parabolic trajectory and x as a function of time is v0 cosine tx0 times t.
00:36 Therefore this time is 75 meters over 31 meters per second and cosine of 40 degree.
00:56 Theta 3 .2 seconds.
01:02 And now we have to find the height at this time...
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