Question

Projectile Motion Problem?

Answer 1

a) `22.46`
b) `15.89`

Explanation:

Supposing the origin of coordinates at the player, the ball describes a parabola such as

`(x,y) = (v_x t, v_y t - 1/2g t^2)`

After `t = t_0 = 3.6` the ball hits the grass.

so `v_x t_0 = s_0 = 50->v_x = s_0/t_0=50/3.6=13.89`

Also

`v_y t_0 - 1/2g t_0^2 = 0` (after `t_0` seconds, the ball hits the grass)

so `v_y = 1/2 g t_0 = 1/2 9.81 xx 3.6 = 17.66`

then `v^2=v_x^2+v_y^2 = 504.71-> v = 22.46`

Using the mechanical energy conservation relationship

`1/2 m v_y^2=m g y_(max)->y_(max) = 1/2 v_y^2/g =1/2 17.66^2/9.81=15.89`

Answer 2

`sf((a))`

`sf(22.5color(white)(x)"m/s"`

`sf((b))`

`sf(15.9color(white)(x)m)`

Explanation:

`sf((a))`

Consider the horizontal component of the motion:

`sf(V_x=Vcostheta=50.0/3.6=13.88color(white)(x)"m/s")`

Since this is perpendicular to the force of gravity, this remains constant.

Consider the vertical component of the motion:

`sf(V_y=Vcos(90-theta)=Vsintheta)`

This is the initial velocity of the ball in the y direction.

If we assume the motion to be symmetrical we can say that when the ball reaches its maximum height `sf(t_(max)=3.6/2=1.8color(white)(x)s)`.

Now we can use:

`sf(v=u+at)`

This becomes:

`sf(0=Vsintheta-9.81xx1.8)`

`:.` `sf(Vsintheta=9.81xx1.8=17.66color(white)(x)"m/s"=V_y)`

Now we know `sf(V_x)` and `sf(V_y)` we can use Pythagoras to get the resultant velocity V . This was the method used in the answer by @Cesereo R.

I did it using some Trig':

`sf((cancel(v)sintheta)/(cancel(v)costheta)=tantheta=17.66/13.88=1.272)`

`sf(theta =tan^(-1)1.272=51.8^@)`

This is the angle of launch.

Since `sf(V_y=Vsintheta)` we get:

`sf(Vsin(51.8)=17.66)`

`:.` `sf(V=17.66/sin(51.8)=17.66/0.785=22.5color(white)(x)"m/s")`

`sf((b))`

To get the height reached we can use:

`sf(s=ut+1/2at^2)`

This becomes:

`sf(s=Vsinthetat-1/2"g"t^2)`

`:.` `sf(s=V_yt-1/2"g"t^2)`

Again, the time taken to reach the maximum height will be 3.6/2 = 1.8 s

`sf(s=17.66xx1.8-1/2xx9.81xx1.8^2)` `sf(m)`

`sf(s=31.788-15.89=15.9color(white)(x)m)`