Projectile Motion ?

enter image source here

The answer is there, but how to solve?

2 Answers
Sep 2, 2017

266color(white)(l)"m" < "distance from shore" < 3469color(white)(l)"m"266lm<distance from shore<3469lm

Explanation:

We're asked to find the distances from the western shore that a ship can be so that it is out of range of the enemy ship's projectiles.

To do this, we can first use the kinematics equation

ul(Deltay = v_0sintheta_0t - 1/2g t^2

where

  • Deltay is the change in height of the projectile

  • v_0 is the initial speed of the projectile (given as 250 "m/s")

  • theta_0 is the launch angle of the projectile (what we'll be finding)

  • t is the time

  • g = 9.81 "m/s"^2

This equation deals with vertical motion; for the horizontal motion, we have the equation

ul(Deltax = v_0costheta_0t

where

  • Deltax is the change in horizontal position of the particle

  • v_0 = 250 "m/s"

  • theta_0 is the launch angle

  • t is the time

One thing we can take note of is that the time when the projectile has a change in height Deltay = 1800 "m" and a change in horizontal position Deltax = 2500 "m" is the same.

Therefore, we can solve the second equation for t and plug that in for t in the first equation, to eliminate the time variable:

t = (Deltax)/(v_0costheta_0)

Deltay = v_0sintheta_0((Deltax)/(v_0costheta_0)) - 1/2g((Deltax)/(v_0costheta_0))^2

Plugging in known values, we have

1800color(white)(l)"m" = (250color(white)(l)"m/s")sintheta_0((2500color(white)(l)"m")/((250color(white)(l)"m/s")costheta_0)) - 1/2(9.81color(white)(l)"m/s"^2)((2500color(white)(l)"m")^2)/((250color(white)(l)"m/s")^2cos^2theta_0)

1800color(white)(l)"m" = (2500color(white)(l)"m/s")tantheta_0 - 1/2(9.81color(white)(l)"m/s"^2)((100color(white)(l)"s"^2)/(cos^2theta_0))

Solving for theta_0 yields

  • theta_0 = ul(50.1^"o" and ul(75.6^"o"

These correspond to the angles theta_L and theta_H in the image.

Now, we go back and use our horizontal distance equation

Deltax = v_0costheta_0t

We still don't know the time, but we can find it using the equation

Deltay = v_0sintheta_0-1/2g t^2

where the change in height Deltay is 0 because we're trying to find the time when it comes back down.

Plugging in known values, we have

0 = (250color(white)(l)"m/s")sin(50.1^"o")t - 1/2(9.81color(white)(l)"m/s"^2)t^2

t = 39.1 "s"

And so for the lower angle theta_L, we have

Deltax = v_0costheta_Lt = (250color(white)(l)"m/s")cos(50.1^"o")(39.1color(white)(l)"s") = color(red)(ul(6269color(white)(l)"m"

The distance from the western shore is thus

"max distance" = color(red)(6269color(white)(l)"m") - (2500color(white)(l)"m" + 300color(white)(l)"m") = color(red)(ulbar(|stackrel(" ")(" "3469color(white)(l)"m"" ")|)

And for the higher angle theta_H = 75.6^"o", the time is

0 = (250color(white)(l)"m/s")sin(75.6^"o")t - 1/2(9.81color(white)(l)"m/s"^2)t^2

t = 49.4 "s"

And so we have

Deltax = v_0costheta_Ht = (250color(white)(l)"m/s")cos(75.6^"o")(49.4color(white)(l)"s") = color(blue)(ul(3066color(white)(l)"m"

The distance from the western shore is

color(blue)(3066color(white)(l)"m") - (2500color(white)(l)"m" + 300color(white)(l)"m") = color(blue)(ulbar(|stackrel(" ")(" "266color(white)(l)"m"" ")|)

Sep 2, 2017

Given that

  • the velocity of projectile u=250m"/"s
  • the minimum vertical height just to cover h=1800m,the height of the mountain peak.
  • the distance of the enemy ship from the mountain base d=2500m
  • taking acceleration due to gravity g=9.8m"/"s^2

Let theta be the angle of projection of projectile required just to fly over the mountain peak to hit the target ship behind the mountain and t sec be the time taken by the projectile to reach at the height of the peak.

Hence the vertical displacement covered by the projectile during this t sec will be

h=usinthetaxxt-1/2g t^2......[1]

And the horizontal displacement during this time will be

d=ucos theta t

=>t=d/(ucostheta).....[2]

Substituting the value of t in [1} we get

h=usinthetaxxd/(ucostheta)-1/2g (d/(ucostheta))^2

=>h=dtantheta-g/2xx(d/u)^2sec^2theta

=>1800=2500tantheta-9.8/2xx(2500/250)^2(1+tan^2theta)

=>490tan^2theta-2500tantheta-2290=0

=>tantheta=(2500pmsqrt(2500^2-4xx490xx2290))/(2xx490)

=>tantheta ~~1.2=>theta=50.2^@

and

=>tantheta ~~3.9=>theta=75.63^@

We know that formula of horizontal range

R=(u^2sin2theta)/g

When theta=50.1^@

The range R_(50.1^@)=(250^2sin(2xx50.1))/9.8~~6276m
For this range the safe distance from western shore should be greater than 6276-2800=3476m

Again when theta=75.63^@

The range R_(75.6^@)=(250^2sin(2xx75.63))/9.8~~3066m
For this range the safe distance from western shore should be less than 3066-2800=266m