Question

2D projectile motion,

Answer 1

35 degrees is the angle of shooting....

iam not sure.. please recheck

Answer 2

hi, did u find the angle first or the initial velocity first?

Answer 3

hi, sorry to disappoint you but, it is not calculated value....

sorry.....

Answer 4

i dont understand, anw, i used this formula vf-vo=at to get my time up there when vf=0. then t-> x2 to get the whole time for the shot til vf became 0 again (reach the ground) then i plugged in in x=vo-1/2at^2 to get my vo value, from there can take that value and use trigonometry to solve for the angle, is this thought process correct?

Answer 5

i'm on it

Answer 6

this is what ive come up the 2nd round, thanks! i wanna check! (:

Answer 7

hey you should apply the formulas for max height of projectile
h=(u^2 x sin^2theta)/2g and equate it with 35m
next you should use the formula for range of the projectile
R=(u^2 x sin^2theta)/g and then equate it by 100
next you divide both the equations and the thing you did is very conceptual and is the best method to do it but sometimes it is lengthy and you've taken some values wrong but i appreciate your try

Answer 8

how do u divide both equations? im kinda confused.

Answer 9

i got the angle 34.99 nearly equal to 35
and the initial velocity
is
14.58

Answer 10

okay
u^2 sin^2theta/2g/u^2 sin2theta/g this is what we get
now by using some property(i dont know the name)
if we write x divided by 1/y , we can write it as x multiplied by y
so (u^2 sin^2 theta)/ 2g divided by u^2 sin2theta/ g equals (u^2 sin^2theta)/2g multiplied by g/(u^2 sin2theta)
and you can write sin 2 theta as 2sin theta cos theta
but if you dont know the formula , i'll upload the correct method that you wanted to approach
please dont mind if this reply is very clumsy

Answer 11

okay i get it already(: thanks alot! :D btw to get the time in the air which formula do we use?

Answer 12

you can use the formula T=(2u sin theta)/g
and you're welcome

Answer 13

okay! thank you! (: