Hi!

Question

i need some help with this ASAP!

a body is projected upwards from a point on level ground with a velocity of 45 m/s at an elevation of 30 degrees. Working from the expression for uniformly accelerated motion in a straight line, determine

1) the time of flight
2) the distance from the point of projection to the point where the body strikes the ground
3) the greatest height reached

THANKS VERY MUCH!@

I will be happy to critiue your work or thinking. POst it, please.

ok

here goes..i want to know if the formula are kool

**** for the time of flight, i did this

= 2vsintheata / g

= 2(45sin30) / 9.81
= 45 / 9.81
= 4.59 s

**** for the distance, i did

= vcostheata * t (someone told me about this formula..is it correct???)

**** for the height, i did

h= ut + 1/2gt^2

u being the velocity which is 45m/s..rite...or 0??

that's my work thus far

Formulas? Formulas are no substitute for thinking.

<<a body is projected upwards from a point on level ground with a velocity of 45 m/s at an elevation of 30 degrees. Working from the expression for uniformly accelerated motion in a straight line, determine

1) the time of flight
2) the distance from the point of projection to the point where the body strikes the ground
3) the greatest height reached>>

Initial vertical velocity is 45m/s*sin30
at the top, the final vertical velcoty is -45m/s*sin30.

Vf= Vi - gt
solve for t, the time of flight.
Then for distance, distance=45*cosTheta*timeflight

heightmax= Vinitialvertical*timeup -1/2 g*timeup^2
where timeup= 1/2 time of flight.

One thinks in words, not formulas, and that is what I am trying to demonstrate. Most of your work is ok, as you will see when you go through my words. Rethink your height calcs.