First slide

Projection Under uniform Acceleration

Question

A body is projected making an angle of 600 with horizontal from the base of an inclined plane having an inclination of 300 with horizontal. When the same body is projected vertically upward with same speed, maximum height attained by it is 45 m. Then its range on the inclined plane is

Difficult

A
60 m
B

80 $5\sqrt{3}$ m
C
30 m
D
25 m

Solution

$\frac{u^{2} \sin^{2} 60^{0}}{2 \times 10} = 45 \Rightarrow u^{2} = 1200 m^{2} / s^{2} T h e r e f o r e r a n g e o n i n c l i n e d p l a n e = \frac{2 u^{2} \sin (α - β)}{g \cos^{2} β} = \frac{2 \times 1200 \times \sin 30^{0}}{10 \times \cos^{2} 30^{0}} m = 80 m$

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