rown horizontally. The landing velocity should make the smallest possible angle, which is $ 30^\circ $, with the horizontal.

Initial Conditions: At $ t = 0 $:

Horizontal velocity component: $ u_x = 10\sqrt{3} \, \text{ms}^{-1} $

Vertical velocity component: $ u_y = 0 \, \text{ms}^{-1} $

At time $ t = t $:

Horizontal velocity remains constant: $ v_x = 10\sqrt{3} $

Vertical velocity due to gravity: $ v_y = g t = 10t $

Condition for Minimum Angle:

The tangent of the angle of the landing velocity with respect to the horizontal is given by:

$ \tan 30^\circ = \frac{v_y}{v_x} $

Substituting the given values:

$ \frac{1}{\sqrt{3}} = \frac{10t}{10\sqrt{3}} $

Solve for $ t $:

$ \frac{1}{\sqrt{3}} = \frac{t}{\sqrt{3}} \Rightarrow t = 1 \, \text{s} $

Calculate Height $ h $:

Using the formula for vertical displacement under gravity:

$ h = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times 1^2 = 5 \, \text{m} $

Thus, the height $ h $ from which the balls should be released so that the landing velocity makes the minimum angle of $ 30^\circ $ with the horizontal is $ 5 \, \text{meters} $.

Motion In A Plane

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