physic buffs, need help with problem

[smokinHawk1647545499]

ok its an almost standard projectile motion problem.

trying to figure it out, hoping that its not some simulanious equation problem or something, the formula to solve alludes me must be some simple way to do it.

 

projectile is launched at an unknown velocity at an 45* angle.

it reaches its target at

Horizonal distance it makes is 70ft.

Verticle distance it makes is 6ft above launch point.

 

I need to find the intial velocity

and time of flight.

 

thanks

[smokinHawk1647545499]

the formula i came up with to find time is

(2*(verticle distance - horizontal distance)/gravity)^(1/2)

 

since its at a 45* angle: velocity horizontal = velocity verticle

 

anyone want to confirm if those are right?

[Berto]

is the target at the same level as the lauch point?

 

your assumption of vertical speed equal to horizontal speed is correct because sin(45) and cos(45) are the same value.

 

the relationship between time of flight and Vo (initial velocity) is easy since you have the constant acceleration of gravity. You can assume that the horizontal velocity remains unchanged because drag effects are negligible (for your class).

 

so then if you know how far the horizontal target is you can get the initial horizontal velocity. and by rule the vertical velocity is the same.

 

 

i might have biffed the explanation, i really like to see drawings of problems :-/

[smokinHawk1647545499]

No the target is 6 feet higher then the launch point

 

something seems to be wrong with my above equation as the numbers arent jiving when i put them back into different equations.

[Trouble Maker]

I'll do it tomorrow, sleep now (physics minor).

[Lustalbert]

ADD keeps me from fully enjoying math like you guys.

[smokinHawk1647545499]

anyone figure this out, i have gotten no where.

 

it cant be that hard, its not like im asking you to define what causes gravity.

[black00ws6]

v sin 45 = vo, substitue v sin 45 for vo in the equation of 2ad = vf^2 - vo^2

 

2 * -9.8 m/s^2 * 1.83 m = (0 m/s)^2 - (v sin 45)^2

-35.87 m^2/s^2 = -(v sin 45)^2

35.87 m^2/s^2 = (v sin 45)^2

sqrt (35.87 m^2/s^2) = v sin 45

5.99 m/s = v sin 45

v = (5.99/sin 45) m/s

v = 8.47 m/s

 

To find the time....

 

 

 

8.47 m/s sin 45 = 5.99 m/s , y component

 

t * 5.99 m/s = 21.34 m

t = 3.56 s

 

I'm sure I f'ed up something.. been about 8 years since doing this crap, jesus, i'm old

[smokin5s]

Jesus John...

[black00ws6]

It's not right, I think i have a conversion off somewhere.. The time and x component should equal the distance traveled.

 

should be right now.... SHOULD be

[smokinHawk1647545499]

thanks

still something looks off, ill go over your figures and check to see if i see something, might be right though i dont know.

[black00ws6]

so was it right?