Question

A projectile is shot from the ground at an angle of `pi/8` and a speed of `2 m/s`. When the projectile is at its maximum height, what will its distance from the starting point be?

Answer 1

`t_m=(2sin(pi/8))/g`

`d=(2sin(pi/8)(4cos^2(pi/8)+sin^2(pi/8)))/g`

Explanation:

`a_y(t)=-g`.

`v_y(t)=inta_y(t)dt=-g t+v_(0y)=-g t+v_0sin(theta)`.

`y(t)=intv_y(t)=-1/2g t^2+v_0sin(theta)t`.

`a_x(t)=0`.

`v_x(t)=inta_x(t)=0+v_(0x)=v_0cos(theta)`.

`x(t)=intv_x(t)=v_0cos(theta)t`.

The projectile is at its maximum height when `v_y(t)=0`, so when `g t=v_0sin(theta)`.

`t_m=(v_0sin(theta))/g=(2sin(pi/8))/g`.

Its `x` distance from the starting point at this time will be `x(t_m)=(v_0cos(theta)*v_0sin(theta))/g=(4sin(pi/8)cos(pi/8))/g`.

Its `y` distance from the starting point at this time will be `y(t_m)=-1/2g((v_0sin(theta))/g)^2+v_0sin(theta)((v_0sin(theta))/g)=(v_0^2sin^2(theta))/(2g)=(2sin^2(pi/8))/g`.

So the total distance `d` from the starting point is given by the Pythagorean Theorem on the distances, i.e. `d^2=x^2+y^2`

So after all the algebra, we get `d=(2sin(pi/8)(4cos^2(pi/8)+sin^2(pi/8)))/g`.

Answer 2

The distance form the starting point, `d ~~ 0.15 m`

Explanation:

The equation for the y coordinate is:

`y(t) = ((-4.9" m")/"s"^2)t^2 + ((2" m")/s")sin(pi/8)t`

The equation for the x coordinate is:

`x(t) = ((2" m")/s")cos(pi/8)t`

The time that the projectile is at is maximum height is at the axis of symmetry of the parabola:

`t = -b/(2a)`

`t = -(2sin(pi/8))/(-9.8)s`

`t ~~ 0.08" s"`

`y(0.08) ~~ 0.03 m`

`x(0.08) ~~ 0.15m`

The distance, d, at the above time is:

`d = sqrt((0.03)^2 + (0.15)^2)`

`d ~~ 0.15 m`