1 a bullet fired from a rifle,attains a maximum height of 25m and crosses a range of 200m find the - Physics - Motion In A Plane

Question

1 a bullet fired from a rifle,attains a maximum height of 25m
and crosses a range of 200m find the angle of projection
2 find the range of ball,which when projected with a velocity of
29 4 m/s just passes over a pole 4 9 m high
3 a projectile is thrown from a point 39 2 m away from the foot
of a building 19 6m high and just reaches the top horizontally find
the velocity of projection and the angle of throw

Jyoti Pant · Accepted Answer

1) We know maximum height

$h = \frac{u^{2} \sin^{2} θ}{2 g} = 25 - - (1) R a n g e R = \frac{u^{2} \sin 2 θ}{g} = 200 - - (2) d i v i d i n g (1) b y (2) a n d s o l v i n g, w e g e t 8 \sin θ = 4 \cos θ \tan θ = 1 / 2 = > θ = {26.6}^{0} 2) u = 29.4 m / s, h = 4.9 m, R = ? w e h a v e h = \frac{u^{2} \sin^{2} θ}{2 g} = > 4.9 = \frac{{29.4}^{2} \times \sin^{2} θ}{2 \times 9.8} = > \sin^{2} θ = \frac{4.9 \times 2 \times 9.8}{{29.4}^{2}} = 1 / 9 \sin θ = 1 / 3 = > θ = {19.47}^{0} n o w R = \frac{u^{2} \sin 2 θ}{g} = \frac{{29.4}^{2} \times 2 \sin θ \cos θ}{9.8} = 55.27 m$

$3) H e r e \max i m u m h e i g h t = 19.6 m h o r i z o n t a l r a n g e = 39.2 + 39.2 = 78.4 m h = \frac{u^{2} \sin^{2} θ}{2 g} = 19.6 - - (1) R = \frac{u^{2} \sin 2 θ}{g} = 78.4 - - (2) D i v i d i n g (1) b y (2) a n d s o l v i n g, w e g e t 4 \sin^{2} θ = 4 \sin θ \cos θ \tan θ = 1 = > θ = 45^{0} n o w p u t t i n g t h i s v a l u e i n (1) \frac{u^{2} \sin^{2} 45}{2 \times 9.8} = 19.6 u^{2} = 19.6 \times 9.8 \times 4 = 768.32 = > u = 27.72 m / s$