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A projectile haswith the horizontal it is fired? (g-9.8 m/s)(g-9.8 m/s.) Ans: (i) 980 m. (11) 10 s, (iii) 138.6 m/sa range of S0 m and it reaches to a maximum height of 10 m. At what angleAns: 38.66 |
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Answer» We know that Horizontal range R = u^2sin 2theta/g --------------- (1) We know that Maximum height H = u^2sin^2 theta/2g ------------- (2) On solving (1) and(2) we get R/H = 4/tan theta --------------------- (3) Given, R = 50m, Height = 10, substitute in (3), we get 50/10 = 4/tan theta tan theta =4/5 theta = 38.6598 degrees. Hope this helps! |
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