1) and 2)
Let's sketch the problem and define the coordinate direction:
3)
We know that the height of the ball is 1.5 meters and its initial velocity is zero.
4)
We can write the following equation for the height of the ball h over the time t:
[tex]\begin{gathered} h=h_0+v_0t+\frac{at^2}{2} \\ h=1.5+0\cdot t+\frac{(-9.8)t^2_{}}{2} \\ h=1.5-4.9t^2 \end{gathered}[/tex]Problem 2.
The acceleration acting on the ball when it goes up is the gravity acceleration.
Its magnitude is equal to 9.8 m/s² and the direction is down (negative y)
The acceleration acting on the ball when it goes down is the same (gravity acceleration: 9.8 m/s² down), since there is no external force acting on the ball.
