The motion of the ball is analyzed by separating it into independent horizontal (x) and vertical (y) components. Let the initial velocity be $\vec{v}_0 = v_{0x} \hat{i} + v_{0y} \hat{j}$. The acceleration is $\vec{a} = -g \hat{j}$, where $g$ is the acceleration due to gravity.

[Q1] What is the initial velocity of the ball when it is thrown?

The horizontal motion occurs at a constant velocity, as there is no horizontal acceleration ($a_x = 0$). The ball travels a distance $s$ in time $t$.

$$s = v_{0x} t \implies v_{0x} = \frac{s}{t}$$

The vertical motion is under constant downward acceleration $a_y = -g$. Since the ball is caught at the same height it was thrown, the total vertical displacement $\Delta y$ is zero. Using the kinematic equation for displacement:

$$\Delta y = v_{0y} t + \frac{1}{2} a_y t^2$$ $$0 = v_{0y} t - \frac{1}{2} g t^2$$

Since $t eq 0$, we can divide by $t$:

$$v_{0y} = \frac{1}{2} g t$$

The initial velocity vector is $\vec{v}_0 = \frac{s}{t} \hat{i} + \frac{gt}{2} \hat{j}$. The initial speed, which is the magnitude of the initial velocity, is found using the Pythagorean theorem.

$$v_0 = |\vec{v}_0| = \sqrt{v_{0x}^2 + v_{0y}^2}$$ $$v_0 = \sqrt{\left(\frac{s}{t}\right)^2 + \left(\frac{gt}{2}\right)^2}$$

[Q2] What is the speed of the ball at its highest point?

At the highest point of its trajectory, the vertical component of the velocity is momentarily zero ($v_y = 0$). The horizontal component of velocity remains constant throughout the flight.

$$v_x = v_{0x} = \frac{s}{t}$$

Therefore, the speed at the highest point is equal to the constant horizontal velocity.

$$v_{\text{highest}} = v_x = \frac{s}{t}$$

[Q3] How high does the ball travel above the point of release?

Let the maximum height be $h$. This is the vertical displacement when the vertical velocity $v_y = 0$. Using the kinematic equation relating velocity, acceleration, and displacement:

$$v_y^2 = v_{0y}^2 + 2 a_y h$$ $$0^2 = \left(\frac{gt}{2}\right)^2 + 2(-g)h$$ $$2gh = \frac{g^2 t^2}{4}$$

Solving for $h$:

$$h = \frac{g^2 t^2}{8g} = \frac{gt^2}{8}$$