Question

Projectile Motion ?

The answer is there, but how to solve?

Answer 1

`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"" ")|)`

Answer 2

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`