Projectile motion after a flying helical path

[qq545282501]

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?

 

[BvU]

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

Always estimate the answer when using a calculator or a spreadsheet.

And I wouldn't round off before I had the answer.

 

[qq545282501]

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

Always estimate the answer when using a calculator or a spreadsheet.

And I wouldn't round off before I had the answer.

so my approach was correct? thank you

 

[BvU]

so my approach was correct? thank you

Approach was correct, execution a bit sloppy. And: You're welcome.