Question

A cubical block of side a is on the ground.A particle is projected from ground with minimum velocity vo such
that it crosses over the cube​

Answer 1

Answer:

Height of projectile at its highest point is h = u^2sin^2Ф/2g

Horizontal distance covered till this point is x= R/2 = u^2sin^2Ф/2g = u^2sinФcosФ/g

Time taken to reach highest point is t= T/2 = u sinФ/g

Hence, displacement is d= √h^2+x^2 = u^2sinФ/2g (√1+3cos^2Ф)

Average velocity v= d/2 = u/2√1+3cos^2Ф

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