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a ball projected with 60 degrees to horizontal u=392 msec find time of flight range anf maximum range |
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Answer» ⭐《ANSWER》 ↪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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