Question

An arrow is thrown in the air its time of flight is 5 sec and the range is 200 m determine the vertical component of velocity of projection

Answer 1

The range of Arrow is 200m

If the arrow has been sent up at an angle thetafrom 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.

u.sin(theta) is the velocity in the vertical direction so if h is the vertical height covered then ,and v is the final velocity which must be zero in the vertical direction and the

time taken to reach this height must be T/2 due to symmetry of upward and downward motion.

v = u.sin(theta)- (g)(T/2) : so , u sin(theta) = 10 x 2.5 sec=25m/s

and h = (1/2) . g. (T/2)^2 = (1/2). 10. (2.5)^2 = (5 x 6.25 )m = 31.25 m




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