Question

A man throws a out a ball horizontally with velocity 6√10m/s and ball strikes at ground making angle 30with horizontal . Find height of building from where ball was thrown

Answer 1

Okay so the ball makes an angle of 30° with horizontal when it hits, so that means velocity makes an angle of 30°.

then tan30° = Vy/Vx

Vy is final velocity in Y direction.Vx in X.
Now X component of velocity will not change during motion, this is a characteristic of projectile motion.
Vx= Ux= 6√10 m/s

tan30° = (1/√3) = Vy/Vx
Vy= (1/√3) Vx
Vy= (1/√3)*(6√10)
Vy= 2√30.

initial velocity in the Y direction was zero!
and acceleration in Y direction is -g (i am taking -10 )

${v}^{2} - {u}^{2} = 2as$
4*30 - 0 = - 2 *10 *s
s= -6 m

negative sign because object moves downwards (I chose down as negative).

Height is simply 6 metres.
Hope it helped