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The trajectory of a projectile in a vert...

The trajectory of a projectile in a vertical plane is `y = ax - bx^2`, where `a and b` are constant and `x and y` are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.

Text Solution

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The correct Answer is:
`(u cos (alpha+theta))/(cos theta)`
Horizontal displacement of particle with respect to ground is zero. It means that initial velocity with respect to ground is only vertical, or there is no horizontal component of the absolute velocity of the particle.
Let v be the velocity of the block down the plane
Velocity of particle `=u cos (alpha+theta)hati+u sin (alpha+theta)hatj`
Velocity of block `=-v cos theta hati-v sin theta hatj therefore " velocity of particle with respect to ground"`
`={u cos (theta+theta)-v cos theta}hati+{u sin (alpha+theta)-v sin theta}hatj`
Now, as we said earlier that horizontal component of absolute velocity should be zero. Therefore, `u cos (alpha+theta)-v cos theta=0 or v=(u cos (alpha+theta))/(cos theta) ("down the plane")`
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