2 added 191 characters in body edited Mar 2 '14 at 1:22 Mister Mystère 51922 silver badges1919 bronze badges 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 $$y(t)=0$$. 1 answered Mar 2 '14 at 1:03 Mister Mystère 51922 silver badges1919 bronze badges 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: $$g=a$$ $$\Rightarrow v(t) = \int^t_0 g.dt=g.t+v_0$$ $$\Rightarrow y(t) = \int^t_0 g.t.dt=\frac{1}{2}g.t^2+v_0.t+y_0$$