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2D Motion as Two 1D Motions

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Question

A ball is projected at an angle of $30^{0}$ above with the horizontal from the top of a tower and strikes the ground in $5$ sec at an angle of $45^{0}$ with the horizontal. Find the height of the tower and the speed with which it was projected.

Height

= 33.5 m

; Speed

= 36.6 m / s

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Height

= 43.6 m

; Speed

= 15 m / s

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Height

= 59.2 m

; Speed

= 30 m / s

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Height

= 71.9 m

; Speed

= 50 m / s

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Solution

The correct option is B Height $= 33.5 m$ ; Speed $= 36.6 m / s$
$L e t i n i t i a l v e l o c i t y a l o n g x d i r e c t i o n b e u_{x} a n d t h a t a l o n g y d i r e c t i o n b e u_{y} \frac{u_{x}}{c o s 30^{0}} = \frac{u_{y}}{s i n 30^{0}} . . . . (1) i n 5 s e c o n d s t h e v e l o c i t y m a k e s a n a n g l e o f 45^{0} w i t h t h e h o r i z o n t a l, w h i c h m e a n s h o r i z n t a l a n d v e r t i c a l v e l o c i t i e s a r e s a m e . v_{y} = u_{y} - g t = - u_{x} = - u_{y} \sqrt{3} \Rightarrow u_{y} = \frac{g t}{1 + \sqrt{3}} = \frac{50}{2.732} = 18.3 m / s u_{x} = u_{y} \sqrt{3} = 31.7 m / s U = \sqrt{u_{x}^{2} + u_{y}^{2}} = 36.6 m / s$

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A ball is projected at an angle of 300 above with the horizontal from the top of a tower and strikes the ground in 5 sec at an angle of 450 with the horizontal. Find the height of the tower and the speed with which it was projected.

A ball is projected at an angle of $30^{0}$ above with the horizontal from the top of a tower and strikes the ground in $5$ sec at an angle of $45^{0}$ with the horizontal. Find the height of the tower and the speed with which it was projected.