Question

Hi everyone, my teacher just gave us this problem and nobody was able to solve it! It's a 2-dimension kinematics problem.
A projectile is fired to the East with an initial velocity of 35m/s and lands 184 m away. What angle was the projectile fired at with respect to the x-axis? Good luck! :o)~

Answer 1

I can tell you that a double angle formula is supposed to be used but after 5 hours trying every combination of trig function from my calc book, I cannot solve this problem!

Answer 2

i know Vx=35cos(theta) and Vy=35sin(theta)

Answer 3

Horizontal range R = u^2 sin 2theta/g
u = initial velocity

Answer 4

deltaX= 184m also Xi = 0 Xf= 184

Answer 5

so 35^2sin2theta/-9.8 is that right?

Answer 6

=184, yes

Answer 7

so i can solve for theta from that?

Answer 8

nopes, something is missing...

Answer 9

sin(2theta)=-1.472

Answer 10

which is not possible

Answer 11

thats what i thought...its like not enough information but she told everyone she wanted to see how sharp we were...i was like whatever lady!

Answer 12

so should i just give up? she claims that we should have used theta/2 as angle under some radical...very confused

Answer 13

idk bout that but i will give one more shot, by finding the time 't'

Answer 14

i even tried spliting the prblem into half by just focusing on the max height and dividing mt range by 2 184/2=92 but that got me nowhere either

Answer 15

I even put 35sintheta for my Viy and 35costheta for my Vix and then equated both of those with time but it still didnt work

Answer 16

i got imaginary time :O
i had to use a= +9.8 to get real time, but that wouldn't be correct

Answer 17

wow...maybe she is crazy

Answer 18

here is a question...even if we made our own hypothetical problem where the angle was 45 degrees, by symmetry, we would know the the final velocity and angle would also be the same, but without either know the time or the distance, we still can't create our own problem in hopes of relating the two can we?

Answer 19

with initial vel as 35 m/s
the maximum dist. the projectile can travel (taking theta=45) will be just 125 m
it would not go 184 m far

distance and time are related as
s=ut+0.5at^2 if thats what u were asking

Answer 20

well thanks for giving it a try!

Answer 21

Maybe it was shot from an elevated place ;)

Answer 22

Hi badhi...how would that have made a difference?

Answer 23

ok...lets say hypothetically, 28m cliff...you could solve it then?

Answer 24

If we write equations, for many \((\theta,x)\) values we can get the required distance - 184. Which will make the question meaningless. But for me it seems to be the only explanation.