A projectile is thrown with velocity of 50 m/sec towards an inclined plane from ground such that it strikes the inclined plane perpendicularly the angle of projection of the projectile is 53 with the horizontal and the inclined plane is inclined at an angle of 45 to the horizontal.find the distance between the point of projection and the foot of inclined plane.
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Let a be angle of projectile and b be angle of the inclined plane.
THE FORMULAE IS :
This is the distance along the plane and it is given by :
S = u Cos (a - b)
u is the initial speed.
Substituting we have :
= 50× Cos (53 - 45)
= 50 × Cos 8 = 49.51 m
THE FORMULAE IS :
This is the distance along the plane and it is given by :
S = u Cos (a - b)
u is the initial speed.
Substituting we have :
= 50× Cos (53 - 45)
= 50 × Cos 8 = 49.51 m
Answered by
2
Answer:
Explanation:
Distance will be 175m. Use vectors and cosine rule.
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