Dynamics Part 3

Solving this you need to abuse the kinematics equation with acceleration: $h = v_0t + \frac{at^2}{2}$ .

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Lets say that $v_1 = 120 (m/s)$ , $t_1 = 3 (s)$ , $h = 100 (m)$ , $v_2 = \boxed{?}$ .

Lets make kinematic equation for the first ball: $h = v_1 t_{100} - \frac{gt_{100}^2}{2}$ .

Plugging in the values and solving for $t_{100}$ we get: $t_{100} \approx 0.8638 or 23.6 (s)$ . However, the first value is way too little for the second ball to catch up (in fact, according to the problem's statement it won't be thrown yet!), so we will use only one value of $t_{100} \approx 23.6 (s)$ .

We need to find such $v_2$ that it would reach $h$ in $\Delta t = t_{100} - t = 20.6 (s)$ .

Constructing the kinematic equation for the second ball: $h = v_2 \Delta t - \frac{g \Delta t^2}{2}$ -> $v_2 = \boxed{\frac{2h + g \Delta t^2}{2\Delta t}}$

Plugging in the numbers: $v_2 = \boxed{105.89}$