if the bullet is fired for maximum horizontal range of 2560m with a speed of 160m/s find the maximum the bullet can reach it (g=10m/s^2)

Question

if the bullet is fired for maximum horizontal range of 2560m  with  a speed of 160m/s  find the maximum the bullet can reach it (g=10m/s^2)

anonymous · Answer

maximum what?

experimentx · Answer

You have maximum range when angle of projection is equal to 45.
horizontal velocity = $ u \cos \theta $
Distance travelled = velocity x time = $ t \times u \cos \theta $ = 2560m

vertical velocity = $ u \times \sin\theta$
since you know time ... find the range

anonymous · Answer

for maximum range, it should be fired at an angle of 45 degrees
$$R = u ^{2}\sin 2\theta /g$$

anonymous · Answer

we have to find the maximum height>>>

ujjwal · Answer

what do you mean by " find the maximum the bullet can reach it " .. I didn't get the question.. Are you sure you haven't made typing errors?

anonymous · Answer

sorry for the mistake

anonymous · Answer

it is maximum height...

ujjwal · Answer

I see.. max height .. wait a moment..

anonymous · Answer

ok

ujjwal · Answer

Maximum horizontal range is for angle of projection 45 degrees.. Now in formula of R put theta = 45 and find u .. Now replace value of u and theta=45 in formula of H.. and you will get it..

anonymous · Answer

we are given the range. so find the angle thete. it comes out to be 45 degrees. now substitute these values in the max. height equation.
$$H = u ^{2} \sin ^{2}\theta/2g$$

anonymous · Answer

@ujjwal  i think we are given the value of initial velocity (u)

anonymous · Answer

yes u=160m/s

anonymous · Answer

go ahead and calculate the max. height

anonymous · Answer

i am getting it as 320  but the answer is 640m

ujjwal · Answer

It is a lot easier than.. you have more info than needed.. go ahead put theta = 45 .. that's all.. and i recommend you to go through the theory part of your text book before you dive into numericals...

anonymous · Answer

a calculation mistake

anonymous · Answer

yea i got theta =30

anonymous · Answer

how u got theta =45 can u show me

anonymous · Answer

$$2560 = (16)^{2}\times \sin2\theta /10$$

anonymous · Answer

i'm sorry its (160)^2

anonymous · Answer

sin 2 theta =1 then what to do???

ujjwal · Answer

And it is a fact that for max range theta is always equal to 45 degrees...

anonymous · Answer

sin 2 theta = sin 90
2 theta = 90
 theta = 45

anonymous · Answer

can u solve for it ?

anonymous · Answer

sin 2 theta = sin 90 how???

anonymous · Answer

sin 90 = 1. so in place of 1, put sin 90

anonymous · Answer

sin90=0

anonymous · Answer

sin 90 = 1 not zero

anonymous · Answer

oh sorry .....

radar · Answer

All you have to do is plug the values that you already have to solve.  Here is the equation$$H=u ^{2}\sin ^{2}\theta /2g$$
$$H=160 ^{2}\sin ^{2}45^{o}/2*10$$
H=(25600)(.5)/20
H=12800/20=640 m