an arrow
is thrown in the air its time of flight is 5 seconds and the range is 200 m determine the vertical component of the velocity of projection the horizontal component and the maximum height and the angle made with the horizontal
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Maximum height reached is 31.25 m.
From the formula of time of flight
2u/g=t (u here is the y component )
2u/10=5
u= 5*10/2
u = 25 m/s
Now applying x to the formula of maximum height
u*u/2g = h
25 * 25/2 *10 = h
31.25 m =h
From the formula of time of flight
2u/g=t (u here is the y component )
2u/10=5
u= 5*10/2
u = 25 m/s
Now applying x to the formula of maximum height
u*u/2g = h
25 * 25/2 *10 = h
31.25 m =h
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