Dear Student, if velocity of the projectile is u then Range =9=u2sin(90)gor u=310m/sand time to reach pole =6ucos(45)=25sso height of the pole =usin(45)t-12gt2=310×12×25-12×10×252=2m Regard

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41 , ttæn,; ?red•, right now at 5 km south of the crossing rs mov•r•-; towards a' 40 closest distance approach between the two Cars be (2) 2.5 km (3) 56 km 40 In a projectile motion x-direction is horizontal y-directlon is vertical. The relation beb.veen y and x is g•ven by y — 10x2. Here x and y are in_ metre. The horizontal range of the prolectile is ( f) 1m 41 A projectile thrown at 450 just crosses the top ot a pole uvmse foot is at 6 m from the point Of projection The projectile falls on the other side at a dstance ot 3 m from the foot of the pole, Height Bf the pola is (3) 28m t A projectile is thrown at 600 wth the horizontal After 4 s It is moving perpendicular to its matial direction of motion. If g = 10 rn}s2, then initial speed of throw d, lic Ion

Dear Student,

i f v e l o c i t y o f t h e p r o j e c t i l e i s u t h e n R a n g e = 9 = \frac{u^{2} \sin (90)}{g} o r u = 3 \sqrt{10} m / s a n d t i m e t o r e a c h p o l e = \frac{6}{u \cos (45)} = \frac{2}{\sqrt{5}} s s o h e i g h t o f t h e p o l e = u \sin (45) t - \frac{1}{2} g t^{2} = 3 \sqrt{10} \times \frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{5}} - \frac{1}{2} \times 10 \times {(\frac{2}{\sqrt{5}})}^{2} = 2 m

Regard

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