The trajectory obtained by eliminating t is y =19-x 2/2.

2. Speed: Vx=2, Vy =-4t；;

Acceleration: ax=0, ay=-4, which is uniform and variable speed.

So the average speed from t= 1.00s to t=2.00s is the progress when t =1.5s. ..

Vx=2，Vy=-6

V = 6.32m m/s

3. t = 1.00 seconds:

Vx=2，Vy=-4

V=4.47

When t = 1s, the slope of the trajectory line y'=-x, i.e.

The tangent direction of the trajectory line at t = 1s is 45 diagonal.

The conclusion is that ax = 0 and ay =-4.

When ay=-4 is decomposed at 45, the tangential and normal acceleration is 4/ 1.4 14=2.828.