Question

In Fig. 4-40, a ball is launched with a velocity of magnitude$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, at an angle of $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{°}$to the horizontal. The launch point is at the base of a ramp of horizontal length ${\mathbf{d}}_{\mathbf{1}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$and heightrole="math" localid="1654153249604" ${\mathbf{d}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{m}$. A plateau is located at the top of the ramp. (a)Does the ball land on the ramp or the plateau? When it lands, what are the (b) magnitude and (c) angle of its displacement from the launch point?

Short answer

(a). The ball lands on the ramp and not on the plateau.

(b). The magnitude of the displacement of the ball from the launch point is $5.82\mathrm{m}$

(c). The angle of its displacement from the launch point is $31°$.

Step-by-step solution

Given information

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

The angle with which the ball is launched$\mathrm{\theta }=50°$

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 ${\mathrm{d}}_2=\mathrm{y}$

Determining the concept of kinematic equation

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

$△\mathrm{y}={\mathrm{xtan\theta }}_0‐\frac{{\mathrm{gx}}^2}{2{\left({\mathrm{v}}_0{\mathrm{cos\theta }}_0\right)}^2}$ (i)

The Newton’s second kinematic equation,

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

Calculating the time and distance travelled

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),

$$\begin{gathered} △\mathrm{y}=\mathrm{xtan}50°‐\frac{9.8x^2}{2{(10\cos 50°)}^2} \\ △\mathrm{y}=0.6\mathrm{m} \end{gathered}$$

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

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

Thus,

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

(a) Determining whether the ball lands on the ramp or the plateau

The horizontal displacement of the ball is,

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_{\mathrm{ox}}\mathrm{t} \\ \mathrm{x}={\mathrm{v}}_{\mathrm{o}}\mathrm{cos\theta }\times \mathrm{t} \\ \mathrm{x}=10\cos 50°\left(0.787\right) \\ \mathrm{x}=4.99\mathrm{m} \end{gathered}$$

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,

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

(b) determining the magnitude of the displacement of the ball from the launch point

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

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

(c) Determining the angle of its displacement from the launch point

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

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

Thus, $\mathrm{\theta }=31°$