Question about velocity/position vectors?

We know the velocity in the x direction is equal to 2.4-1.6t^2. Since position is the integral of the velocity, the x position is described by 2.4t-(1.6/3)t^3+c and since we know the x position at t=0 is 0, we know c is zero so x=2.4t-(1.6/3)t^3. Setting it to zero yields t=0 (already known) and 2.4-(1.6/3)t^2=0 so t^2=7.2/1.6 and t=sqrt(7.2/1.6) (positive in this case since the situation after t=0 is required). Since our y velocity is 4t, the y-position is 2t^2+c with c being 0 again so y=2t^2. Filling in the t we found before, we get y=2*(7.2/1.6)=14.2/1.6 which is the required y-position.

The position vector r is given by the integral of the velocity.

r = (2.4t - 1.6t^3/3)i + 2t^2 j

It x coordinate is zero when (2.4t - 1.6t^3/3) = 0

You can take it from here to solve for t and plug that in for the j component of r to find the altitude.

vertical velocity = dy/dt = 4t

vertical position, y = ∫4tdt = 2t² + y(0) = 2t² since we are given that y(0) = 0

horizontal velocity = dx/dt = 2.4-1.6t²

horizontal position = x = ∫(2.4-1.6t²)dt = 2.4t - (16/30)t³ since we are given that x(0) = 0

Now the question is to evaluate y(t) at x = 0 and t > 0

x= 0 = t(2.4-16t/30)

t= 0

16t/30 = 2.4

t = 2.4*30/16 = 4.5

Therefore the bird crosses the y-axis (x=0) at t=0 and t=4.5. So what is the bird's altitude at t = 4.5

y = 2t² = 2(4.5)² = 40.5