- #1

mohlam12

- 154

- 0

a rocket was launched 33degrees above the horizontal at a speed of 760m/s. it accelerates 15m/ss until it burns out after 10sec. how many sec did it say in the air?

here s what i did:

vertical speed = 760sin33 = 413.9 m/s

horizontal velocity = 637.4 m/s

so from the vertucal speed, i calculated the time that stays in the air, which is 413.9/9.8= 42.2 sec

that s the time that took the ricket to go up, so i multiplied by 2, and finally, the time that stayed in the air is 84.4

BUT, i didn t use the acceleration that happened during 10sec ! is what i did above right??

thank you and i appreciate ur help! :)

mohammed