Question

A ball has an initial velocity of 1.30 m/s along the +y axis and, starting at t0, receives an acceleration of 2.10 m/s^2 in the +x direction.

(a) What is the position of the ball 2.50 s after t0?
(b) What is the velocity of the ball at that time?

Any help is appreciated.

Answer 1

a)
There is no acceleration in x direction so coordinate x can be found from this equation:
\[x=v_0t\]but there is acceleration in y direction with no initial velocity in this direction so y coordinate is simply:\[y=\frac{ at^2 }{ 2}\]
b)
the velocity remains the same in x direction:
\[v_x=v_0\]but changes in y direction(due to acceleration)\[v_y=at\]total velocity is then \[v=\sqrt{v_x^2+v_y^2}\](Pythagoras theorem)