Question

If I have a 0.0565 kg ball and I have to shoot it 10m at 10m/s and im shooting it at 45 degrees, how high should I make my launcher? (preferably less than 1.5m)
Thanks

Answer 1

what do you mean by "i have to shoot it at 10m"? Are you mentioning the target height or range there? For the values that you mentioned, the maximum height reachable is 2.55m, so you need a 7.45m high launcher to reach the target height, which is not less than 1.5.
If you mentioned the target range as 10m, then the maximum range achievable with the values you have is 10.2, so you may not need a launcher.
If this answer is wrong, I guess I couldn't understand your question, sorry..

Answer 2

yes @mruna_cp is right, if the total horizontal distance to be achieved is 10m , the height of the launcher will have to be negative. that is below the ground level.


let the relative height reached(from the launcher) = h
initial velocity is u(10m/s) at 45 deg.

let us resolve this velocity into its horizontal and vertical components
\[Ux = Ucos45\] - This remains constant throught the journey of the projectile
\[Uy = Usin45\] - This changes as there is acceleration due to gravity in the vertical direction

Now, let us divide the total time of the projectile's journey into two parts.
t1 = time taken for upwards jouney
t2 = time taken for downward journey

so for t1, during the upward journey
consider the vertical velocity at the maximum height reached. its 0 there.
so \[v = u - g*t1 \] where v = 0, u = Ucos45
\[t1 = Ucos45/g\]

the maximum height reached is
\[m*g*h = 1/2*m*Uy^2\] where Uy is the initial vertical velocity
using energy conservation. Note that horizontal velocity remains constant.

hence we get \[h = Uy^2/2g\]

now in downward journey, projectile will go some extra height downwards which is the height of the launcher.
let the height of launcher be h1
hence total height H = h + h1
\[hence applying , S = u*t + 1*a*t^2\]
S = H, u = 0, a = g
\[H = 1*g*t^2\]
hence,
\[t = \sqrt{2*H/g}\]
Thus, t2 = \[\sqrt{2*(H+h)/g}\]

so total time = t1 + t2
we know horizontal velcity = Ucos45
hence total horizontal distance covered by projectile =
d = Ucos45 * (t1 + t2)

so if d = 10,
\[10 = U/\sqrt{2} * (U/\sqrt{2} g+ \sqrt{2*(H+h)/g})\]

solve this to find h.

we get on solving,
h = g - 10
so unless g > 10 which doesn't sit right for earth's surface, this would be a negative height for the launching platform, meaning it would have to be below ground level.... so start digging ;)

Answer 3

Is there any way that it could be possitive, like I know i must sound really dumb but the velocity is tentative, so that can change

Answer 4

first of all, we are all quite dumb before the mysteries of the universe, so do not worry :)
so yes, if you decrease your initial velocity, then you can achieve a positive height of the platform.
think of it this way.
if you throw something at an angle from the first floor quite fast it falls a certain distance away on the road.
Now if you throw it gently, you could still achieve the same range if say you throw it from the 2nd floor.

Answer 5

but like how high would i have to make it?

Answer 6

ready my solution above.
in my solution above, calculate using smaller values of u. then you will keep getting increasing heights.