Question

a body is projected with a velocity U at an angle Alpha() with horizontal. the velocity of the body will become perpendicular to the velocity of the projector after a time T which is given by???

Answer 1

Answer on Question # 76089, Physics - Mechanics - Relativity:

Question: A body is projected with a velocity $U$ at an angle Alpha ($\alpha$) with horizontal. The velocity of the body will become perpendicular to the velocity of the projector after a time $T$ which is given by?

Solution: A body is projected with initial velocity $U$ by making angle $\alpha$ with the horizontal. Then after time $T$ (consider at point $P$) it's direction is perpendicular to $U$.

Magnitude of velocity at this point (P) is given by $v = U \cot \alpha$

Vertical component of initial velocity $v_i = U \sin \alpha$

Final velocity at point P is $(v_f) = -v \cos \alpha = -U \cot \alpha \cos \alpha$

Time of flight up to point P is T.

Now we know, $v_f = v_i - gT$ (considering vertical motion)

Or, $U \cot \alpha \cos \alpha = U \sin \alpha - gT$ [Plug $v_f$ and $v_i$ value]

$$\mathrm{Or}, \mathrm{T} = \frac{\mathrm{U} \sin \alpha + \mathrm{U} \cot \alpha \cos \alpha}{\mathrm{g}} = \frac{\mathrm{U} \cosec \alpha}{\mathrm{g}}$$

Answer: $T = \frac{U \cosec \alpha}{g}$

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