# Projectile motion after a flying helical path

## Homework Statement

A bird flew along a helical path $$r(t)=<5cost,5sint,2t>$$, at time t=10 second, the bird died instantly. where did it hit the xy-plane ? g=32 ft/s/s.

## Homework Equations

$$r(t)=(v_0 \cos\theta t) i+(v_0 \sin\theta t - 0.5gt^2 + h_0) j$$

## The Attempt at a Solution

taking derivative of r(t), I got the velocity function respect to t, $$v(t)=<-5sint,5cost,2>$$
at t=10 s, we are located at (-4.2, -2.7, 20)
so that means the velocity in the x-direction at time =10 seconds is 2.72 ft/s
and the velocity in the y-direction at t=10 s = -4.2 ft/s
to hit the ground means that the distance in Z direction is 0, setting $$0=z(10)+v_z(10) t - 16t^2$$
velocity in z direction is constant= 2
we can find the time the bird hit the ground, which = 1.18 second.
now, plug t and velocity in the x-direction into $$X_f=x(10)+v_x(10) t$$, we can find x-axis distance=-9.9 ft
$$Y_f=y(10)+v_y(10) t$$ similarly, we can find y - coordinate= -76.56 ft ( this value seems way off)

i am confused, does angle of projection matters in this case? if does, how do we find the angle of projection?

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$X_f=x(10)+v_x(10) t = -4.2 + 2.7 * 1.18$ can never be -9.9 ft !
For y you simply have a decimal point error, I'd guess: $\ -2.7 - 1.18 * 4.2 = -7.7$

$X_f=x(10)+v_x(10) t = -4.2 + 2.7 * 1.18$ can never be -9.9 ft !
For y you simply have a decimal point error, I'd guess: $\ -2.7 - 1.18 * 4.2 = -7.7$