Question

An object is launched at a velocity of 33 m/s in a direction making an angle 34 degrees upward with the horizontal. What is the total flight time of the object ? (assume g = 9.81 m/s2)

Answer 1

Vy Vi Un Initial velocity 0; = 33 m/s Angle O= 34" Aceleration due to gravity g-9.8 ingin men Ke vertical Horizontal Uy; = Visimo = Vicolo ay=-g In vertical direction object reachey its manimun height it time 'ť and velocity velocity becomes aerofaccelerating due to gravity toward downwardy] Valia an0 Usy =0 m/s fy = y + ata V. g ascending o= Visino-gxta Vi sino ta 33x sim 340 - 1.883sec 9.8 Time of ta= 1.883 sec T= tarta td is time of descending which is equal to ta -a ta = 2x1.883 - 3.766 sec of flight Time of flight Ta ime T-3.966 sec 000) Uy; 33x Sin 34' mis Vytomis ) UR wy - qah Yf 19.6 o-- (33x singur) --9x9.8xh Manimum height he (33 xSingu) 17.374 h= lyfta + hata time of descen cending from heighth 17.374 = oxta + **9.8x dd 2x17-324 -3.546 td is 9.8 -1.883 sec tä ( Ed ime of flight T= taxta of flight taxta = 3766sec