Question

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at what angle should a body be projectile when a velocity 24 MS inverse just to pass over the obstacle 16 m high at a horizontal distance of 30 to take ji equals to 10 metre second square inverse ​

Answer 1

If point of projection is taken as the origin of the co-ordinate system the projected body must pass through a point having coordinates (32m,16m).

If u be the initial velocity of the projectile and theta the angle of projection then

Horizontal component of initial velocity,

ux= u cos theta

Vertical component of initial velocity, uy= u sin theta

If the body passes through point P after time T then horizontal distance covered ,

$x = (u \cos(theta) )t$

or

$32 = (24 \cos(theta) )t$

Similarly, vertical distances covered,

$y = (u \sin(theta)) t - \frac{1}{2} gt ^{2}$

or

$16 = (24 \sin(theta)) t - \frac{1}{2} \times 10 \times t ^{2}$

From equation (1),

$t = \frac{32}{24 \cos(theta) }$

Putting this value of t in equation (2), we get

$16=(24cos theta ) 32/24 cos theta - 1/2 × 10× (32/24cos theta)^2$

or

$16=32 tan theta -5×16/9 cos^2 theta$

or

$1=2 tan theta - 5/9 sec^2 theta$

or

$5 tan^2 theta -18 tan theta +14=0$

Therefore,

$tan theta = 18+ root (18)^2-4×5×14/10$

or

$tan theta = 18- root(18)^2-4×5×14/10$

=2.462 or 1.137

Hence theta =67°54' or 48°40'