initial velocity?

Question

how would you find the initial velocity of an projectile, if you are only given the angle, the height, and the width?

Answer 1

Not sure what you mean by width but angle and height will be enough.

If you were lifted to the max height position, and dropped the projectile, it would hit the ground with a speed equal to the vertical component of the velocity you need to find. (Potential energy converted into kinetic.) To get the speed when hitting the ground after being dropped, use the formula
V^2 = Vo + 2*g*height
where Vo = 0 (It had 0 speed before you dropped it)

So you have the vertical component of its initial velocity. You have the angle, so use trig to get the resultant.

Answer 2

Weight (or mass) of the object is irrelevant since we don't need to find force, energy, or momentum.

Gravitational attraction can be assumed -9.81 m/s^2 (or -32.2 ft/s^2) for near-surface terrestrial objects.

Wind resistance can also be neglected for a first approximation - especially if other data is unknown like size and surface area of the object under investigation.

Answer 3

Without the weight, gravitational and wind resistance, along with what caused it to go in the 1st place and how powerful that was, I don't see how you can.

Chance of giving a crap + effort - complaining X lethargic / thinking = answer

Answer 4

x(t)=e^(-t){-2cos(3t) - (two/three)sin(t)} The preliminary role is while t = zero: x(zero) = a million { -two*a million - two/three * zero } x(zero) = - two feet. <==== Initial pace = dx/dt evaluated at t= zero: dx/dt = - e^(-t){-2cos(3t) - (two/three)sin(t)} + e^(-t){6sin(3t) - (two/three)cos(t)} At t = zero dx/dt = - a million { -two*a million - two/three * zero} + a million {6*zero - two/three * a million)} ....... = two - two/three dx/dt = eleven/three feet/s <==== Ans. Hope this is helping