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A projectile has a range of 50m and Reaches a maximum of 10m. what is the elevation of the projectile? |
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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. |
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