A vertical pole has red mark at some height A stone is projected from fixed point on the ground wehen projected at angle of 45 degree it hits the pool at 90 degrees 1 metre above the mark. when projected with different speed at angle of tan inverse 3/4 it hit the pool at 90 degrees 1.5 metre below the mark . Find speed and angle of projection so that it hit the mark orthogonal to the pole
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First see pic 2, then pic 2
Final Answer : angle of Projection is such that
tan (a) = 9/10
Initial Speed of projectile = 20.02m/s
Steps and Understanding :
1) To find the speed and angle of a projection, we need
=> Maximum Height
=> &Range
2) Here,
Point on ground is constant
or Point from where stone is projected in both cases is same.
=> Range is same in both cases.
Hence, we will make equations in terms of Maximum Height and Range from given equations.
3) u1 and u2 are the initial speed of projectile in given cases.
H = Maximum Height
2R = Range
For Calculation see pic.
Final Answer : angle of Projection is such that
tan (a) = 9/10
Initial Speed of projectile = 20.02m/s
Steps and Understanding :
1) To find the speed and angle of a projection, we need
=> Maximum Height
=> &Range
2) Here,
Point on ground is constant
or Point from where stone is projected in both cases is same.
=> Range is same in both cases.
Hence, we will make equations in terms of Maximum Height and Range from given equations.
3) u1 and u2 are the initial speed of projectile in given cases.
H = Maximum Height
2R = Range
For Calculation see pic.
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