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Initially, angle made by horizontal = (90 - 60) = 30°

When the projectile is making 90° with the initial, then angle made by the projectile with horizontal = 60°.

Since there is no acceleration in horizontal, horizontal velocity component will remain same:

ucos(30) = vcos(60)

=> v = 100

Considering vertical components of velocities in the equation v = u + at:

=> -vsin60 = usin30 + (-10)t

and using v = 100, we get the time** t = 20 secs**

( this is not matchng with the answers given, so I guess the options given might be misprinted or it could be the question. Although, if we take the initial angle to be 60° **with the horizontal, **the we get ans t = 20/ )