Question

A projectile is fired horizontally with a velocity u. Show that its trajectory is a parabola. Also obtain expression for (i) time of flight (ii) horizontal range (iii) velocity at any instant.

Answer 1

please find your answer in the image.

Answer 2

Answer:

Suppose a body is projected up with an initial velocity u in a direction making an angle θ with the horizontal. Resolving u in horizontal and vertical components, we have

$u_x = u\hspace{1mm} cos\theta\\u_y = u\hspace{1mm} sin\theta$

(i) Time of flight:

Suppose the body takes time t to reach the highest point P of its path. The vertical velocity of the body at P is 0. ∴ in equation v = u - gt,

$v = v_y = 0$

and $u = u_y = u\hspace{1mm} sin\theta$

$0 = u\hspace{1mm}sin\theta - gt$

$t = \frac{u\hspace{1mm}sin\theta}{g}$

Beyond point P  the body moves downward along its parabolic path. It takes the same time t in moving from P to O. Hence, time of flight is given by

$T = 2t = \frac{2u\hspace{1mm} sin \theta}{g}$

(ii) Horizontal Range:

The body covers a horizontal distance OO' during its flight. This distance is called the 'horizontal range' R. It is given by

R = uₓ × T

    = u cos θ × $\frac{2u\hspace{1mm} sin \theta}{g}$

$R = \frac{u^2\hspace{1mm} sin 2\theta}{g}$

(iii) Velocity at any instant

Velocity at any instant is given by

$\overrightarrow v(t) = v_x\hat i + v_y\hat j\\where\hspace{1mm}v_x\hspace{1mm}is \hspace{1mm}the \hspace{1mm}horizontal\hspace{1mm}component\hspace{1mm}of \hspace{1mm}v \hspace{1mm}and\hspace{1mm}v_y\hspace{1mm}is \hspace{1mm}the\hspace{1mm}vertical\hspace{1mm}component\hspace{1mm}of\hspace{1mm}v.$

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