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Equations of Motion

A projectile ...

Question

A projectile is thrown in the upward direction making an angle of $60^{o}$ with the horizontal direction with a velocity of $147$ $m s^{- 1}$ . Then the time after which its inclination with the horizontal is $45^{o}$ , is?

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Solution

Vy /Vx =tan45 or Vy =Vx

Vx remains constant =147cos60

Vy now using equation of motion for verticle direction
Vy=Uy-gt
Uy=147sin60
Uy=147cos60
putting in eq 1
 t=5.36 sec

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Q. A projectile is thrown in the upward direction making an angle of

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with the horizontal direction with a velocity of

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Then the time after which its inclination with the horizontal is

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Q. a projectile is thrown in the upward direction making an angle of 60 degree with the horizontal direction with a velocity of 147 metre per second then the time after which its inclination with the horizontal becomes 45 degree is ????

My thinking:
Uy= 147sin(60) and. Vy= 147sin(45)
Using Vy=Uy-gt
What is wrong in my thinking. ?????

Q. A projectile is thrown in the vertically upward direction making an angle of 60 with the horizontal direction with a velocity of 150 m/s. the time after which its inclination with the horizontal is 45, is

A)15(√3-1)s B)15(√3+1)s C) 7.5 (√3-1)s D)7.5(√3+1)s

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