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You haven't defined[Edit: If you don't consider air friction and if you're not asked anything about the range of the projectile, but I am going to assume the problem isequations are the same as a vertical freefallfreefall]. If you are trying to find the velocity of the object at any given time, it is not $v_iy$$v_{iy}$ that you need to calculate since it is the initial velocity of the object at $t=0$.

Step by step for a vertical freefall (1D) with origin on the ground at the vertical of the initial position of the object for $t=0$ and y axis toward the groundobject: $$g=a$$$$-g=a_y$$ $$\Rightarrow v(t) = \int^t_0 g.dt=g.t+v_0$$$$\Rightarrow v(t) = \int^t_0 -g.dt=-g.t+v_{0y}$$ $$\Rightarrow y(t) = \int^t_0 g.t.dt=\frac{1}{2}g.t^2+v_0.t+y_0$$$$\Rightarrow y(t) = \int^t_0 (-g.t+v_{0y}).dt=-\frac{1}{2}g.t^2+v_{0y}.t+y_0$$ Here $v_{0y}$ is 0 and $y_0=80m$. You're interested in $v(T_g/2)$ where $T_g$ is $t$ so that $y(t)=0$.

You haven't defined anything, but I am going to assume the problem is a vertical freefall. If you are trying to find the velocity of the object at any given time, it is not $v_iy$ that you need to calculate since it is the initial velocity of the object at $t=0$.

Step by step for a vertical freefall (1D) with origin at the position of the object for $t=0$ and y axis toward the ground: $$g=a$$ $$\Rightarrow v(t) = \int^t_0 g.dt=g.t+v_0$$ $$\Rightarrow y(t) = \int^t_0 g.t.dt=\frac{1}{2}g.t^2+v_0.t+y_0$$

[Edit: If you don't consider air friction and if you're not asked anything about the range of the projectile, the equations are the same as a vertical freefall]. If you are trying to find the velocity of the object at any given time, it is not $v_{iy}$ that you need to calculate since it is the initial velocity of the object at $t=0$.

Step by step for a vertical freefall (1D) with origin on the ground at the vertical of the initial position of the object and y axis toward the object: $$-g=a_y$$ $$\Rightarrow v(t) = \int^t_0 -g.dt=-g.t+v_{0y}$$ $$\Rightarrow y(t) = \int^t_0 (-g.t+v_{0y}).dt=-\frac{1}{2}g.t^2+v_{0y}.t+y_0$$ Here $v_{0y}$ is 0 and $y_0=80m$. You're interested in $v(T_g/2)$ where $T_g$ is $t$ so that $y(t)=0$.

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You haven't defined anything, but I am going to assume the problem is a vertical freefall. If you are trying to find the velocity of the object at any given time, it is not $v_iy$ that you need to calculate since it is the initial velocity of the object at $t=0$.

Step by step for a vertical freefall (1D) with origin at the position of the object for $t=0$ and y axis toward the ground: $$g=a$$ $$\Rightarrow v(t) = \int^t_0 g.dt=g.t+v_0$$ $$\Rightarrow y(t) = \int^t_0 g.t.dt=\frac{1}{2}g.t^2+v_0.t+y_0$$