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A body is projected from the top of a tower with a velocitybar(u) = 3 hat(i) + 4 hat(j) + 5 hat(k) ms^(-1),"where " hat (i), hat(j) and hat(k)are unit vectors along east, north and vertically upwards respectively. If the height of the tower is 0 m , horizontal range of the body on the ground is (g = 10 ms^(-2)) |
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Answer» 15 m u = 5 m/s `hat(k)`is GIVEN vertically upward direction) `a = - 10 m//s^(2)` h = - 30 m Now using the formula `h = ut +(1)/(2) at^(2)` `- 30 = 5t - (1)/(2) xx 10 xx t^(2)` `rArr t^(2) - t - 6 = 0 ` `rArr t = - 2 s `(Not possible) or t = 3 s In this time, PROJECTILE moving in EAST and North with speeds 3m/s and 4m/s . Distances covered in these directions are , In east (x- coordinate) = speed `xx` time = `3 xx3 = 9 m ` and in North (y- coordinate) `= 4 xx 3 = 12 m ` So, Projective lando at (x,y) = (9m,12m) mark So, horizontal range of body on ground is : Range = `SQRT(9^(2)+12^(2))` `= sqrt(225)` 15 m  
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