If a rocket is launched at a 30 degree angle, with a velocity of 36.7 m/s, an average time of 4.3025 s, and a distance of 116.63 m, what is the highest height it reaches in the air?
Average time of what exactly? And what's that distance supposed to be? Your question doesn't make any sense - please restate it more accurately.
Average time the rocket is in the air. The distance is how far the rocket traveled. I need delta y max, or the height at the highest point the rocket reaches while in the air.
So your rocket is not actually a rocket but is ahm.. "thrown" ??? Or do you have numbers on how much fuel is burned and what the exhaust velocity is etc? Distance on the ground?
Yes, distance on the ground. And yes, sorry, the rocket is launched from a tennis ball launcher, so thrown is correct. This was an activity we actually did in physics class, so these are real numbers.
Okay, just for the future - not just here online - try to be really specific and precise, when it comes to ahm.. science and stuff ^_^ Also: Where exactly are you stuck at? Have you gotten anywhere on your own, yet?
Yeah, sorry about that, I guess I forgot about those details. I mainly need an equation that applies to this situation. Or is it using trig? That's mostly it--I'm not really sure what I should be using to solve it.
Well ahm.. there's really more than one way to do it. For starters: Do you think that - in general - the time an object stays in the air has anything to do with how far it moves horizontally?
Well...we got an equation [delta x = V (sub x) * delta t] where delta x was the distance covered and delta t was the change in time. That's what I assumed it was talking about.
Well ahm.. that equation doesn't really have to do with what you want to know at all.. so ahm.. about my previous question - what do you think? |dw:1354261442965:dw| If you were to throw 2 objects like the poor drawing I made. Would one object be longer in the air if they both reach the same height? (we are disregarding friction here, but that's ok)
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