A rubber ball is dropped onto a ramp that is tilted at 20°, as shown in FIGURE P4.59. A bouncing ball obeys the “law of reflection,” which says that the ball leaves the surface at the same angle it approached the surface. The ball’s next bounce is 3.0 m to the right of its first bounce. What is the ball’s rebound speed on its first bounce?

The inclination of the ramp from the ground is θ = 20⁰
The distance between the first and the second bounce of the ball is d = 3m

Firstly, draw the vector component of the velocity of the ball

Now, to understand the motion of the ball, the figure can be seen from another perspective, such that the ramp is horizontal to the ground as shown in the figure,

The horizontal velocity of the ball along the ramp = v cos70⁰
The vertical velocity of the ball =v sin70⁰
The horizontal component of the gravity = g sin20⁰ = g cos70⁰
The vertical component of the gravity = g cos20⁰ = g sin70⁰
Now, let the distance between the two bounces is x
From the figure shown above
$x=\frac{3}{\cos 20°}......\left(1\right)$
Using the equation of motion, for the vertical range,
$$\begin{gathered} y=v_{y}t-\frac{1}{2}{g}_y{t}^2 \\ 0=v\sin 70°t-\frac{1}{2}g\sin 70°t^2 \\ v\sin 70°t=\frac{1}{2}g\sin 70°t^2 \\ t=\frac{2v}{g}......\left(2\right) \end{gathered}$$
Now, using the equation of motion to determine the horizontal range, substitute the value of t in the given equation;

role="math" localid="1650538819547" $$\begin{gathered} x=v_{x}t+\frac{1}{2}{g}_x{t}^2 \\ x=v\sin 70°\left(\frac{2v}{g}\right)+\frac{1}{2}g\sin 70°{\left(\frac{2v}{g}\right)}^2 \\ x=\frac{2v^2\sin 70°}{g}+\frac{2v^2\sin 70°}{g} \\ x=\frac{4v^2\sin 70°}{g}.......\left(3\right) \end{gathered}$$

Equate the equation (1) and (3)