91影视

The initial velocity of ball ${\mathrm{v}}_0=10\mathrm{m}/\mathrm{s}$

The angle with which the ball is launched$\mathrm{胃}=50掳$

The horizontal length of the base of the ramp${\mathrm{d}}_1=6\mathrm{m}$

The height of the base of the ramp${\mathrm{d}}_2=3.6\mathrm{m}$

Consider here x and are the horizontal and vertical displacement. Thus${\mathrm{d}}_1=\mathrm{x}$and ${\mathrm{d}}_2=\mathrm{y}$

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Also this problem deals with the projectile path. When a body is projected with velocity making a certain angle with horizontal, it follows the parabolic trajectory known as the projectile.

Using these equations and equation for the projectile path, whether the ball lands on the ramp or the plateau, the magnitude of the displacement of the ball and the angle of its displacement from the launch point can be found.

Formula:

The equation for projectile path

$鈻�\mathrm{y}={\mathrm{虫迟补苍胃}}_0鈥�\frac{{\mathrm{gx}}^2}{2{\left({\mathrm{v}}_0{\mathrm{肠辞蝉胃}}_0\right)}^2}$ (i)

The Newton鈥檚 second kinematic equation,

$\mathrm{y}=\left({\mathrm{v}}_{\mathrm{oy}}\right)\left(\mathrm{t}\right)+\frac{1}{2}{\mathrm{at}}^2$ (ii)

The horizontal displacement of the ball is $\mathrm{x}=6\mathrm{m}.$

The vertical displacement of the ball is $\mathrm{y}=3.6\mathrm{m}.$

Using equation (i),

$$\begin{gathered} 鈻�\mathrm{y}=\mathrm{xtan}50掳鈥�\frac{9.8x^2}{2{(10\cos 50掳)}^2} \\ 鈻�\mathrm{y}=0.6\mathrm{m} \end{gathered}$$

Now, using equation (ii), the vertical displacement of the ball is,

$$\begin{gathered} \mathrm{y}=\left({\mathrm{v}}_{\mathrm{oy}}\right)\left(\mathrm{t}\right)鈥�\frac{1}{2}{\mathrm{gt}}^2 \\ 3.6=0鈥�\frac{1}{2}{\mathrm{gt}}^2 \end{gathered}$$

Thus,

$\mathrm{t}=0.787\mathrm{s}$

The horizontal displacement of the ball is,

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_{\mathrm{ox}}\mathrm{t} \\ \mathrm{x}={\mathrm{v}}_{\mathrm{o}}\mathrm{肠辞蝉胃}脳\mathrm{t} \\ \mathrm{x}=10\cos 50掳\left(0.787\right) \\ \mathrm{x}=4.99\mathrm{m} \end{gathered}$$

As x is less than ${\mathrm{d}}_1$, so the ball does land on the ramp and not on the plateau.

The vertical displacement of the ball is,

$$\begin{gathered} \mathrm{y}=0.6脳=0.6脳4.99 \\ \mathrm{y}=2.99\mathrm{m} \end{gathered}$$

The magnitude of the displacement of the ball from the launch point is,

$$\begin{gathered} \mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2} \\ \mathrm{r}=\sqrt{4.{99}^2+2.{99}^2} \\ \mathrm{r}=5.817 \\ ~5.82\mathrm{m} \end{gathered}$$

The angle of displacement of the ball from the launch point is,

$$\begin{gathered} \mathrm{胃}={\tan }^{鈥�1}\frac{\mathrm{y}}{\mathrm{x}} \\ \mathrm{胃}={\tan }^{鈥�1}\left(\frac{2.99}{4.99}\right) \end{gathered}$$

Thus, $\mathrm{胃}=31掳$