Projectile Motion ?
The answer is there, but how to solve?
The answer is there, but how to solve?
2 Answers
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
Therefore, we can solve the second equation for
#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 = ul(50.1^"o"# and#ul(75.6^"o"#
These correspond to the angles
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
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
#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
#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"" ")|)#
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
Hence the vertical displacement covered by the projectile during this
And the horizontal displacement during this time will be
Substituting the value of t in [1} we get
and
We know that formula of horizontal range
When
The range
For this range the safe distance from western shore should be greater than
Again when
The range
For this range the safe distance from western shore should be less than