A projectile launched upwards from ground has a range of 180m and its timeof flight is 10 s.The velocity(in m/s) after start is (angle of projection with horizontal is theta)?

Question

anonymous · Answer

Do u know the formulae for range and time of flight?

anonymous · Answer

yes

anonymous · Answer

after doing all that i am unable to get the answer......the answer should be 18

anonymous · Answer

simple
$$t=\frac{2u \sin \theta}{g}=10$$
range R
$$R=\frac{u^2 \sin 2\theta}{g}$$
use these

anonymous · Answer

@NotSObright  i used this and not getting

anonymous · Answer

@NotSObright  y got it.....

anonymous · Answer

Wait you are looking for both the angle and the velocity?

anonymous · Answer

no i want velocity after 5s

anonymous · Answer

also
$$u \cos \theta *t$$
derivable from here that 
$$u \cos \theta=18$$

anonymous · Answer

Well I don't think you can do it with time. http://en.wikipedia.org/wiki/Trajectory_of_a_projectile look down to where it says velocity at x not at time t find the distance traveled in 5s and plug that to the equation

anonymous · Answer

the distance after 5 seconds would be 90m or half of the total distance

anonymous · Answer

ok thanzz  now i get it....

anonymous · Answer

@Brent0423  thanzz

anonymous · Answer

okay great! let me know what you come up with for the answer

anonymous · Answer

18m/s

anonymous · Answer

is that correct

anonymous · Answer

If you were not provided a given starting angle then yes that answer is correct

anonymous · Answer

oh....no the angle is given as theta

anonymous · Answer

@Brent0423  is there any way

anonymous · Answer

@Brent0423  r u there

anonymous · Answer

do you know the degrees of theta? It would really help if you could draw out the problem so i can see the entire problem in order to provide you with a correct answer

anonymous · Answer

no

anonymous · Answer

??

anonymous · Answer

no angle

anonymous · Answer

can you still draw out the problem please