An arrow is shot into the air its range is 200ms and time of flight is 5 sec if the value of g is 10m/s then the horizontal component of the velocity of arrow is
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The range of Arrow is 200m
If the arrow has been sent up at an angle theta from horizontal with initial velocity u;
then in 5 secs flight time it has travelled in horizontal direction with uniform velocity u. cos(theta)
Therefore 200m = u.cos(theta) x T ; u cos(theta) = 200/5 = 40 m/s
the arrow has to go to a height h as its decelerated in vertical direction and has to return to the ground in T sec.
If the arrow has been sent up at an angle theta from horizontal with initial velocity u;
then in 5 secs flight time it has travelled in horizontal direction with uniform velocity u. cos(theta)
Therefore 200m = u.cos(theta) x T ; u cos(theta) = 200/5 = 40 m/s
the arrow has to go to a height h as its decelerated in vertical direction and has to return to the ground in T sec.
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