Refer to the diagram shown below.
The following discussion assumes a simplistic analysis that ignores air resistance and variations in the terrain that the missile travels over.
Let the launch velocity be V₀ at an angle of θ relative to the horizontal.
The horizontal component of velocity is V₀ cosθ.
If the time of flight is
![t_{f}](https://tex.z-dn.net/?f=t_%7Bf%7D)
, then
![r=V_{o} \, t_{f}](https://tex.z-dn.net/?f=r%3DV_%7Bo%7D%20%5C%2C%20t_%7Bf%7D)
where r = the range of the missile.
Also, the time, t, when the missile is at ground level is given by
![0=V_{o} sin\theta \, t- \frac{1}{2}gt^{2}](https://tex.z-dn.net/?f=0%3DV_%7Bo%7D%20sin%5Ctheta%20%5C%2C%20t-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D%20)
where g = acceleration due to gravity.
t = 0 corresponds to when the missile is launched. Therefore
![t_{f} = \frac{2V_{o}sin\theta}{g}](https://tex.z-dn.net/?f=t_%7Bf%7D%20%3D%20%20%5Cfrac%7B2V_%7Bo%7Dsin%5Ctheta%7D%7Bg%7D%20)
Therefore
![r= \frac{2V_{o}^{2} sin\theta cos\theta}{g} = \frac{V_{o}^{2} sin(2\theta)}{g}](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7B2V_%7Bo%7D%5E%7B2%7D%20sin%5Ctheta%20cos%5Ctheta%7D%7Bg%7D%20%3D%20%5Cfrac%7BV_%7Bo%7D%5E%7B2%7D%20sin%282%5Ctheta%29%7D%7Bg%7D%20)
Typically, θ=45° to achieve maximum range, so that
![r= \frac{V_{o}^{2}}{g}](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7BV_%7Bo%7D%5E%7B2%7D%7D%7Bg%7D%20)
This analysis is more applicable to a scud missile rather than a powered, guided missile.
Answer:
![t_{f} = \frac{r}{V_{o} cos\theta} \\\\ r= \frac{V_{o}^{2} sin(2\theta)}{g}](https://tex.z-dn.net/?f=t_%7Bf%7D%20%3D%20%20%5Cfrac%7Br%7D%7BV_%7Bo%7D%20cos%5Ctheta%7D%20%5C%5C%5C%5C%20r%3D%20%5Cfrac%7BV_%7Bo%7D%5E%7B2%7D%20sin%282%5Ctheta%29%7D%7Bg%7D%20%20)
Usually, θ=45°
Answer:
for abt 36 seconds
Explanation: cuz thats how long it was
coulomb's law. inverse square law of e static attraction and/or repulsion
Answer:
0.152 m
Explanation:
The condition for constructive interference is
d sinθ= mλ (m= 0,1,2,3...)
the slit width
d= 1/N
d= 10^(-3)/520
= 1.92×10^(-6)
The angular spread for the red light is
![\theta_r= sin^{-1}(\frac{m\lambda}{d})](https://tex.z-dn.net/?f=%5Ctheta_r%3D%20sin%5E%7B-1%7D%28%5Cfrac%7Bm%5Clambda%7D%7Bd%7D%29)
![\theta_r= sin^{-1}(\frac{656\times10^{-9}}{1.92\times10^{-6}})](https://tex.z-dn.net/?f=%5Ctheta_r%3D%20sin%5E%7B-1%7D%28%5Cfrac%7B656%5Ctimes10%5E%7B-9%7D%7D%7B1.92%5Ctimes10%5E%7B-6%7D%7D%29)
= 19.97°
The angular spread of blue light is
![\theta_b= sin^{-1}(\frac{m\lambda}{d})](https://tex.z-dn.net/?f=%5Ctheta_b%3D%20sin%5E%7B-1%7D%28%5Cfrac%7Bm%5Clambda%7D%7Bd%7D%29)
![\theta_b= sin^{-1}(\frac{456\times10^{-9}}{1.92\times10^{-6}})](https://tex.z-dn.net/?f=%5Ctheta_b%3D%20sin%5E%7B-1%7D%28%5Cfrac%7B456%5Ctimes10%5E%7B-9%7D%7D%7B1.92%5Ctimes10%5E%7B-6%7D%7D%29)
=14.66°
The distance between the first order red fringe and blue fringe is,
y= y_r- y_b = Ltanθ_r -Ltanθ_b
=1.50( tan19.97°-tan14.66°) = 0.152 m
Given that,
Angle = 30°
Initial velocity = 15 m/s
We need to calculate the time of flight
Using formula of time of flight
![T=\dfrac{2u\sin\theta}{g}](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7B2u%5Csin%5Ctheta%7D%7Bg%7D)
Where, u = initial velocity
g = acceleration due to gravity
Put the value into the formula
![T=\dfrac{2\times15\sin30}{9.8}](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7B2%5Ctimes15%5Csin30%7D%7B9.8%7D)
![T=1.5\ sec](https://tex.z-dn.net/?f=T%3D1.5%5C%20sec)
We need to calculate the final velocity of the ball
Using equation of motion
![v=u+gt](https://tex.z-dn.net/?f=v%3Du%2Bgt)
![v=15+9.8\times1.5](https://tex.z-dn.net/?f=v%3D15%2B9.8%5Ctimes1.5)
![v=29.7\ m/s](https://tex.z-dn.net/?f=v%3D29.7%5C%20m%2Fs)
Hence, The final velocity of the ball is 29.7 m/s.