Question

An Arrow is short in air its time of flght is 5sec and horizontal range is 200 m.the inclination if the arrow with the horizontal is .

Answer 1

1. R = Xo * T = 200 m. 

Xo*5 = 200 

Xo = 40 m/s. = Hor. component of initial 

velocity. 

2. Tr = T/2 = 5 / 2 = 2.5 s. 

Y = Yo + gt. 

Yo = Y - gt = 0 - (-10)*2.5 = 25 m/s. = 

Ver. component of initial velocity. 

hmax = (Y^2-Yo^2)/2g. 

hmax = (0-(25)^2) / -19.6 = 31.9 m.

Answer 2

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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