91影视

The inclination of the ramp from the ground is 胃 = 20鈦�
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 cos70鈦�
The vertical velocity of the ball =v sin70鈦�
The horizontal component of the gravity = g sin20鈦� = g cos70鈦�
The vertical component of the gravity = g cos20鈦� = g sin70鈦�
Now, let the distance between the two bounces is x
From the figure shown above
$x=\frac{3}{\cos 20掳}......\left(1\right)$
Using the equation of motion, for the vertical range,
$$\begin{gathered} y=v_{y}t-\frac{1}{2}{g}_y{t}^2 \\ 0=v\sin 70掳t-\frac{1}{2}g\sin 70掳t^2 \\ v\sin 70掳t=\frac{1}{2}g\sin 70掳t^2 \\ t=\frac{2v}{g}......\left(2\right) \end{gathered}$$
Now, using the equation of motion to determine the horizontal range, substitute the value of t in the given equation;

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

Equate the equation (1) and (3)