1.

a ball projected with 60 degrees to horizontal u=392 msec find time of flight range anf maximum range​

Answer»

⭐《ANSWER》

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↪Actually welcome to the concept of the PROJECTILE MOTION !

↪basically we KNOW that , in the case of the projectiles, the motion PATH or the TRAJECTORY is a Parabolic Function .

↪Also we know that ,

↪The time of flight in the projectile Motion is given as ,

〽T = 2u Sin ⊙ / g

Now here , we given as , U = 392 m/s , ⊙ = 60°

so substituting we get as ,

T = 2 ( 392) Sin 60° / 10

so we get as ,

T = 2*root3/2 *(392) ÷ 10

T = 392*ROOT 3/10 = 67.8

〽⭐so here , the time of flight is , T = 67.8 seconds

Now also here ,

↪RANGE is given as ,

〽R = u^2 Sin2⊙ / g

so we get as ,

R = (392)^2 *1/2 ÷ 10

〽⭐R = 76832/10 = 7683.2 mtrs

↪Now here , the MAXIMUM HEIGHT is given as ,

〽H = u^2 sin^2 ⊙ / 2g

so we get as ,

H = (392)^2 Sin^2 (60) / 2 (10)

H = 153664 * (root3/2)^2 / 20

H = 153664 * (3/4) /20

H = 115248 /20

〽⭐H = 5762.4 mtrs



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