Question

The initial velocity of a body thrown at an angle to the horizon is \[V_0\]The maximum range is S.
At what angle \[\alpha \]to the horizon should the body be thrown to have a range equal to \[l \space ? \space (l 13 years ago

Answer 1

So you just want an angle that is less than S? To know this, you should know that 45degrees is the maximum angle for any launch.

Answer 2

This is nt the answer
& i think here this approach isn't workable.

Answer 3

& one more thing @Coda 45 degrees is nt the maximum angle
It is the angle at which the range is maximum.

Answer 4

You want a range that is less than the maximum range. (l<s) Knowing the maximum range is achieved by 45degrees. What does that tell you?

Answer 5

I don't know but the answer of this question in my book is given as: -\[\LARGE{\color{red}{\alpha = \frac{1}{2} \sin^{-1} \frac{gl}{V_0^2}.}}\]

Answer 6

Alright, where do you think you start using the kinematics / mechanics?

Answer 7

wt do u mean?

Answer 8

Sorry, I meant to say what is it exactly you need help with? I can draw a diagram and we can go from there if you would like.

Answer 9

ya i would like but how to get that answer which i gave from my book!

Answer 10

Work out range for fixed initial speed, then solve for angle.
There should be 2 solutions for each value of range less than maximum range.

Answer 11

u mean like this.
\[S=\frac{V_0^2 \sin2( 45^o) }{g}=\frac{V_0^2}{g}\]
\[l<S.\]Therefore \[l=\frac{V_0^2 \sin2 \alpha}{g}<\frac{V_0^2}{g}\]
But now wt to do @Vincent-Lyon.Fr sir ?

Answer 12

Actually I am nt getting the answer:(

Answer 13

You have worked out:

\(l=\Large \frac{V_0^2 \sin2\alpha }{g}\), which means
\(\sin2\alpha=\Large \frac {gl}{V_0^2}\)

for \(l<\Large \frac {V_0^2}{g}\) the sine has two solutions:

\(2\alpha=\sin^{-1}\Large \frac {gl}{V_0^2}\) or \(2\alpha=\pi-\sin^{-1}\Large \frac {gl}{V_0^2}\) which leads to:

\(\alpha_1=\Large \frac12\sin^{-1}\Large \frac {gl}{V_0^2}\) or \(\alpha_2=\Large \frac\pi2- \frac12\sin^{-1}\Large \frac {gl}{V_0^2}\)

Shot for \(\alpha_1\) is called 'straight shot', as opposed to \(\alpha_2\). I do not know the English term for it, but the French call it 'bell shot'. It might be the same in English.

Answer 14

oh thanx a lot sir:)

Answer 15

gt it ^_^