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"#

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#