ANSWER
7.95 m/s
EXPLANATION
The least initial velocity is the one for which when the object reaches the height of the platform its velocity is zero. From the velocity equation we have,
[tex]v=v_o-gt[/tex]
If v = 0,
[tex]v_o=gt[/tex]
To find the velocity we have to find the time. From the displacement equation,
[tex]y=v_ot-\frac{1}{2}gt^2[/tex]
Replace v0 by the expression above,
[tex]\begin{gathered} y=gt^2-\frac{1}{2}gt^2 \\ y=\frac{1}{2}gt^2 \end{gathered}[/tex]
We know that the height of the platform is 3.2m. Solving this equation for t,
[tex]t=\sqrt[]{\frac{2y}{g}}=\sqrt[]{\frac{2\cdot3.2m}{9.81m/s^2}}\approx0.81s[/tex]
If the object is in the air for 0.81 seconds before reaching the platform, its initial velocity is,
[tex]v_o=gt=9.81m/s^2\cdot0.81s=7.95m/s[/tex]
The least initial velocity needed from ground level for the object to reach the platform is 7.95 m/s
