Question

A projectile is shot from the ground at an angle of `pi/12` and a speed of `7 /18 m/s`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

distance from starting point is `5/1296m~~3.858"x"10^-3m`

Explanation:

Range, `R=V^2/gsin(2theta)`

When the particle reaches its maximum height, the distance from starting point will be `R/2`

`R/2=V^2/(2g)sin(2theta)` where V is the initial launching speed

`R/2=(7/18)^2/(2g)sin(2*pi/12)`

`R/2=(49/324)/(2g)sin(pi/6)`

[Take `g=9.8ms^-2`]

`R/2=(5/648)sin(pi/6)`

`R/2=5/1296m~~3.858"x"10^-3m`

Answer 2

`"distance" = 0.00389` `"m"`

Explanation:

We're asked to find the distance from the launch point a projectile is when it reaches its maximum height.

We can use the equation

`(v_y)^2 = (v_(0y))^2 - 2gh`

to find the maximum height `h`.

When the particle is at its maximum height, the instantaneous velocity `v_y` is `0`.

The initial `y`-velocity `v_(0y)` is

`v_(0y) = v_0sinalpha_0 = (7/18color(white)(l)"m/s")sin(pi/12) = 0.101` `"m/s"`

Plugging in known values, we have

`0 = (0.101color(white)(l)"m/s")^2 - 2(9.81color(white)(l)"m/s"^2)h`

`h = ((0.101color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)) = color(red)(5.16xx10^-4` `color(red)("m"`

To find the horizontal distance covered `Deltax`, we use the equation

`Deltax = v_(0x)t`

`v_(0x) = v_0cosalpha_0 = 0.376` `"m/s"`

The time `t` when it reaches its maximum height is given by

`v_y = v_(0y) -g t`

Rearranging this equation and plugging in known values:

`t = (v_(0y)-v_y)/g = (0.101color(white)(l)"m/s" - 0)/(9.81color(white)(l)"m/s"^2) = 0.0103` `"s"`

Thus, we have

`Deltax = (0.376color(white)(l)"m/s")(0.0103color(white)(l)"s") = color(green)(0.00385` `color(green)("m"`

The distance from the starting point is thus

`r = sqrt((Deltax)^2 + h^2) = sqrt((color(green)(0.00385color(white)(l)"m"))^2 + (color(red)(5.16xx10^-4color(white)(l)"m"))^2)`

`= color(blue)(0.00389` `color(blue)("m"`

Answer 3

THe distance is `=0.0039m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=7/18*sin(1/12pi)`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=7/18sin(1/12pi)-g*t`

`t=7/(18g)*sin(1/12pi)`

`=0.0103s`

The greatest height is

`h=(7/18sin(1/12pi))^2/(2g)=0.00052m`

Resolving in the horizontal direction `rarr^+`

To find the horizontal distance, we apply the equation of motion

`s=u_x*t`

`=7/18cos(1/12pi)*0.0103`

`=0.00387m`

The distance from the starting point is

`d=sqrt(h^2+s^2)`

`=sqrt(0.00052^2+0.00387^2)`

`=0.0039m`