Question

78. A ball is thrown with velocity u making an angle teta with the horizental . Its velocity vector is normal
to initial veocity (u) vector her a time interval of
​

Answer 1

horizontal component: ucostheta

vertical component : usintheta

angle between them : theta = √(ux)²+(uy)²

point where angel is 90°

=> horizontal remains same

=> vertical component Is -vcostheta

angle between = theta = √(ux)²+(vy)²

vertical component would be : -v²cos²theta

.°. v=u+at

=> -vcostheta = ucostheta - gt

=> (v+u)costheta = gt

=> (v+u)costheta/g = time

thus , at this time angle is 90°

Answer 2

Answer:

Step-by-step explanation:

if a ball is thrown with velocity u making an angle theta with the horizontal

then time t is  $\frac{2u sin\alpha }{g}$

The velocity vector is perpendicular  to initial velocity at the very highest point when time is $\frac{2 u sin\alpha }{2g}$

"2" cancels out so the answer will be$\frac{u sin\alpha }{g}$

hope you understood

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