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Q.

An object is thrown towards the tower which is at a horizontal distance of 50 m with an initial velocity of 10 ms-1 and making an angle 30° with the horizontal. The object hits the tower at certain height. The height from the bottom of the tower, where the object hits the tower is (Take, g = 10 ms-2)       

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a

5031-103m

b

5031-103m

c

10031-103m

d

10031-103m

answer is A.

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Detailed Solution

Consider the diagram

Question Image

For horizontal motion, 50 = 10 cos 30° x t

  t=5cos30°=532=103

For vertical motion,

h=10sin30°×103-12×10×1032

    =10×12×103-1003×53=100312-53

    =5031-103m

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