the length of the side opposite side from the angle divided by the length of the side of the longest side (the hypotenuse) is the sine of the angle in a right angle. Try that and verify it is correct.
if horizontal flight time is t=2xVxcos(45)/a
distance horizontally is txVxcos(45)
combining them is 2xv^2*cos^2(45)/a and for
100fps initial velocity on level ground, this means a distance of 312.5 feet
Note that (from above) v^2sin(2x45)/g also works and gives 312,5 ft
If this correct, thanks...
Sure t=2Vy/a - thus 2x70.7/32=4.41875 secs
which then gives 312ft for horizontal distance
from 4.41875x70.7. This all from a muzzle velocity of 100 fps
aimed at a 42degree angle
is that correct?
so for muzzle velocity Vm, I solve for t as indicated and get Vm/16=t.
Then multiply that t by VmxCos(45) or Vm x.707; so it seems I simply
multiply muzzle velocity by .707 to get the distance for any muzzle velocity.
Is it that simple?
If I shoot a ball out of a cannon with muzzle velocity of 100 fps at 45 degree angle, how do I compute how far I shoot it? Ignore air resistance for this.
with the help of many and if I assume I jump down from 2ft and come to a stop from final velocity to 0 over the course of 0.1 second:
given D=1/2at^2
by solving for t: t=(2d/a)^0.5
and given Vf=at (final velocity just before impact)
given that my G factor is given by: (Vf/0.1)/a
by...
Assume I have shoes on and jump down onto concrete, how do you compute the
g-force. Assume I weight 70kg if that matters. G(accelerationDueToGravity)=9.8m/sec;
I think it should be a simple equation, i.e. you start at v=0 and then hit v=final
which is when you hit the concrete. So...
I jump off a high bench (5ft high) to the ground. What is the g-force of the impact and how do you calculate it. It seems mass is irrelevant for this
Joe