ANSWER
7.95 m/s
Step-by-step 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,
![v=v_o-gt](https://img.qammunity.org/2023/formulas/physics/high-school/iej1l2noz1l3uwxknn7xp8v5nbi9g44k75.png)
If v = 0,
![v_o=gt](https://img.qammunity.org/2023/formulas/physics/high-school/1i7ekd3fu7lp6ieajtjg0coylymb11y0mn.png)
To find the velocity we have to find the time. From the displacement equation,
![y=v_ot-(1)/(2)gt^2](https://img.qammunity.org/2023/formulas/physics/high-school/a9ogs00d12axytcq5lp7waahigxz0bsb6b.png)
Replace v0 by the expression above,
![\begin{gathered} y=gt^2-(1)/(2)gt^2 \\ y=(1)/(2)gt^2 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/shukw4dlwyonhl34v7u2r4isxa14c3aguf.png)
We know that the height of the platform is 3.2m. Solving this equation for t,
![t=\sqrt[]{(2y)/(g)}=\sqrt[]{(2\cdot3.2m)/(9.81m/s^2)}\approx0.81s](https://img.qammunity.org/2023/formulas/physics/high-school/v5mypn5jff829lg8vl28p2onjoi2dgtnmi.png)
If the object is in the air for 0.81 seconds before reaching the platform, its initial velocity is,
![v_o=gt=9.81m/s^2\cdot0.81s=7.95m/s](https://img.qammunity.org/2023/formulas/physics/high-school/2qmymlk53amm9kc7yevienmcp2b7svl2rc.png)
The least initial velocity needed from ground level for the object to reach the platform is 7.95 m/s