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Question

A projectile is fired with an initial speed of 5 m/s at an angle of 45with the horizontal, up an inclined plane having angle of inclination 30 from horizontal. What is the range of this projectile?
(Take g=10 m/s2)

A
0.5 m
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B
1.05 m
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C
2.05 m
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D
2.5 m
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Solution

The correct option is B 1.05 m
α= angle of projection from horizontal=45
β=angle of the inclined plane from horizontal=30
The range of a projectile on an inclined plane is given by:

R=2u2sin(αβ)cosαgcosβR=2×52×sin(4530)×cos45g×cos30

Applying formula of sin(AB)=sinAcosBcosAsinB

R=50×(sin45×cos30cos45×sin30)×1210×32R=1.05 m

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