The point from where a ball is projected is taken as the origin of the coordinates axes the X and Y components of it's displacement are given by x =6t andy=8t -5t *t what is the velocity and angle of projection with horizontal ?
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Answered by
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X=6t
【t=x/6】--------1
Put the value of t in y
【Y=8x/6 - 5x^2/36】-------2
Or equation of trajectory..
Now, the time of flight at final point wher y=0.
Put y=0 in eq y=8t-5t^2
U will get, 【t=8/5 secs】
Now, Range will be also at (final point, 0)
So put y=0 in eq 2
U will get,
X or range=48/5
And range = u^2sin2theta/g
....
.
.
.
.
Do maths yourself
.. put variables in formula of time of flight u will het usintheta =8
Now put usintheta in formula of range u will get ucosteta =6
Now if u know how to calculate velocity than do it on your own..
..
.
U=10 m/sec
Hope it helped
【t=x/6】--------1
Put the value of t in y
【Y=8x/6 - 5x^2/36】-------2
Or equation of trajectory..
Now, the time of flight at final point wher y=0.
Put y=0 in eq y=8t-5t^2
U will get, 【t=8/5 secs】
Now, Range will be also at (final point, 0)
So put y=0 in eq 2
U will get,
X or range=48/5
And range = u^2sin2theta/g
....
.
.
.
.
Do maths yourself
.. put variables in formula of time of flight u will het usintheta =8
Now put usintheta in formula of range u will get ucosteta =6
Now if u know how to calculate velocity than do it on your own..
..
.
U=10 m/sec
Hope it helped
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