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Two projectiles are thrown in the air with the same initial speed. one was launched at 20° while the other was launched at 70°. Both projectiles landed at the same point. Did they stay in the air for the same amount of time ?

Sagot :

Answer:

No, projectile launched at 20° landed first.

Explanation:

70° projectile had a vertical velocity of v*sin(70), which is significantly bigger than v*sin(20). Since v=vx+at, we can see that [tex]Vyi*sin(20)-gt < Vyi*sin(70)-gt[/tex]

solve for t, you can check that by substituting velocity for any number.

banwarikath

Answer:

well,they may or they may not. It all depends on the vertical components of the speed with which the projectile is thrown.

Explanation:

to derive the time of flight :

Considering the vertical motion (along y axis) , the journey from the ground to highest point

v=u+at ,where v=final velocity , u=initial velocity , acceleration due to gravity i.e g and t is the time taken by the particle to reach the highest point

o=u(sin theta) + (- g)t

t=u(sin theta) - g

Hence total time of flight is twice of that i.e

T=2(vertical components of speed) / g

So, if two projectile have some vertical components of velocity , they will have equal time of flight.

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