Problem 8.52

8.52. a)

From conservation of momentum,

m₁ v₁  =  (m₁ + m₂) v_f

solving for v_f, we use the conservation of energy, namely

$\frac{1}{2} (m_{1}+m_{2}) v_{f}^{2} ~=~ (m_{1} + m_{2}) g h$

which gives us

$v_{f} = \sqrt{2gh}$

Then, solving for v₁, we find that

$v_{1} ~=~ \frac{m_{1}+m_{2}}{m_{1}} \sqrt{2gh}$

v₁  =  6.30 m/s

b)

In part b, we are basically solving a kinematics problem like we solved in chapter 2. The initial velocity is only in the $\hat{x}$ direction and the acceleration is only in the $\hat{y}$ direction, and it is due to gravity. We find

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

and in the $\hat{x}$ direction

x = v_i t + 0

solving one of these equations for time and then substituting that value in to the other equation we find

$v_{i}^{2} ~=~ \frac{g x^{2}}{2 y}$

which we can rearrange to get into the form that they are looking for

$v_{i} ~=~ \frac{x}{\sqrt{2y/g}}$

plugging in numbers we find that the velocity is

v_i  =  6.17 m/s