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)

Question

Infinity Learn · Accepted Answer

Consider the diagramFor horizontal motion, $$50$$ $$=$$ $$10$$ $$cos$$ $$30$$° x t⇒  $$t=5cos30$$°$$=532=103For$$ vertical motion,$$h=10sin30$$°×$$103-12$$×$$10$$×$$1032$$    $$=10$$×$$12$$×$$103-1003$$×$$53=100312-53$$    $$=5031-103m$$

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