(a) सिद्ध कीजिए कि किसी प्रक्षेप्य के x - अक्ष तथा उसके वेग के बिच के कोण को समय के फलन के रूप में निम्न प्रकार से व्यक्त कर सकते हैं
` theta (t) = tan ^(-1) ((v_(0y) - gt)/(v_(ox)))`
(d) सिद्ध कीजिए कि मूल बिंदु से फेंके गए कोण का मान ` theta_0 = tan ^(-1) ((4h_(m))/(R)) ` होगा | यहाँ प्रयुक्त प्रतीकों के अर्थ समान्य हैं |

Explanation

Let a projectile is projected with velocity `vecv` in a direction making an angle `theta_(0)` with horizontal .
The component of velocity `vecv` are `v_(ax)=v cos theta_(0)` along x-axis and `v_(oy)=v sin theta_(0)` along vertically upwards direction.
Let the projectile be at point P after time. then at P projectile has two velocities.

(i) Uniform velocity `v_(ax)` along `vec(PA)` (horizontally) and
(ii) `(v_(oy)-gt)` along `vec(PB)` (vertically up)
`therefore` Resu ltant velocity at P is
`v=sqrt((v_(ax))^(2)+(v_(oy)-gt)^(2))` in magnitude and direction of velocity at P is given by
`tan theta=(AC)/(PA)=(PB)/(PA)=(v_(oy)-gt)/(v_(ax))`
`theta=tan^(-1)[(v_(oy)-gt)/(v_(ax))]`
(b) Maximum height `h_(m)=(u^(2)sin^(2)theta_(0))/(2g)`
and horizontal range, `R=(u^(2)sin2theta_(0))/(2g)`
and horizontal range, `R=(u^(2)sin 2theta_(0))/(g)`
Dividing, we get,
`(h_(m))/(R)=(u^(2)sin^(2)theta_(0))/(2g)xx(g)/(u^(2)sin2theta_(0))`
`=(sin theta_(0))/(4 cos theta_(0))=(1)/(r) tan theta_(0)`
`tan theta_(0)=(4h_(m))/(R)`
`theta_(0)=tan^(-1){(4h_(m))/(R)}`