A projectile is thrown in the upward direction making an angle of 60 with the horizontal direction with a velocity of 147 m/s. Then what is the time after which its inclination with the horizontal is 45 ?

Question

vikas askiitian expert · Answer

let after time t its inclination with horizontal is 45

at this time Vy /Vx =tan45 or Vy =Vx

Vx remains constant =147cos60 =Vy

now using equation of motion for verticle direction

Vy=Uy-gt..........eq1

Uy=147sin60 ,Uy=147cos60

putting in eq 1,

t=5.36 sec

Shriya · Answer

Suppose the body is projected with the velocity U at an angle of 60 degree with the horizontal and after time t its velocity becomes we make a angle of 45 degree with the horizontal. equating the horizontal and vertical components of U and v, Ucoscos 60 degree equals to b cos 45 degree or b equals to you cos 60 degree upon cos 45 degree Also we sin 45 degree equals to sin 60 degree minus theta from the equation of motion You cos 60 degree upon cos 45 degree into sin 45 degree equals to you sin 60 degree minus theta t equals to you upon g sin 60 degree minus cos 60 degree equals to 1 4 7 upon 9.8 into root 3 by 2 minus one by two Therefore time equals to 5.49 seconds.

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A projectile is thrown in the upward direction making an angle of 60 with the horizontal direction with a velocity of 147 m/s. Then what is the time after which its inclination with the horizontal is 45 ?

A projectile is thrown in the upward direction making an angle of 60 with the horizontal direction with a velocity of 147 m/s. Then what is the time after which its inclination with the horizontal is 45 ?

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A projectile is thrown in the upward direction making an angle of 60 with the horizontal direction with a velocity of 147 m/s. Then what is the time after which its inclination with the horizontal is 45 ?

A projectile is thrown in the upward direction making an angle of 60 with the horizontal direction with a velocity of 147 m/s. Then what is the time after which its inclination with the horizontal is 45 ?

2 Answers

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