A building 75m high has a roof of square shape. A ball is thrown from a point on ground at a speed of 80/√3m/s at an angle 60° with horizontal in the vertical plane which includes the diagonal of the roof of building,grazing the corners of the roof.(g=10)
1.) The distances of point of projection from the building is?
2.)The time for which the ball is above the roof?
3.) The length of one edge of the roof of building?

The equation of path of projectile is given by
y=tanθ·x-g2v02cos2θ.x2
Now we have v0=803m/s, g=10 m/s2, θ=60°, putting all this values we get
y=tan60°x-102×8032×cos260°x2y=3x-3320x2
Now to graze corner the height of the particle will be 75m, so here we have y=75m so we have 75=3x-3320x2x=2003,403.
1.) The distance from the point of projection to the building is 403=69.282m.

2.) now the difference between this two point is 2003-403=803. And the horizontal component of the velocity of the particle is 803×cos60°=403m/s . So the time taken is 803403=2s.

3) Now if a is the side of the roof then diagonal is given by a2. Now we have
a2=803a=806=32.66m.
So the length of the edge of the roof is 32.66m.
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