91影视

• Initial velocity of the ball, $V_0=33.53\mathrm{m}/\mathrm{s}$.
• Angle of the projectile, $胃=55.0^0$.
• Height of wall, $h=11.28\mathrm{m}$.
• Time taken to clear the wall, $t=5.00\mathrm{s}$.

When the particle is launched with some velocity from the ground, its motion is called projectile motion. The path of the particle is calculated by using the set of kinematic equations. For the calculations, the acceleration in the horizontal direction is assumed to be zero, and the upward direction is assumed to be positive. The vertical acceleration is the gravitational acceleration.

Formulae:

The first kinematic equation of motion,$V_f=V_0+at$ (i)

The second kinematic equation of motion,$鈭�x=V_{0}t+\left(\frac{1}{2}\right)脳a脳t^2$ (ii)

The third kinematic equation of motion,${V_f}^2={V_0}^2+2脳a脳鈭�x$ (iii)

Here, $V_0$is the initial velocity,$v_f$ is the final velocity,$a$ is acceleration,$t$ is time, and $鈭�x$is the displacement.

We know that horizontal motion is with constant speed, so$a_x=0$

Components of initial velocity:

$$\begin{gathered} V_{0x}=V_0脳\cos \left(胃\right) \\ =33.53脳\cos \left(55掳\right) \\ =19.23\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} V_{0y}=V_0脳\cos \left(胃\right) \\ =33.53脳\sin \left(55掳\right) \\ =27.47\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence horizontal distance covered in 5.0 s is given using equation (ii). Substitute the values in the equation (ii) to find the values of x-components.

$$\begin{gathered} x=19.23脳5+0 \\ =96.15\mathrm{m} \\ 鈮�96.2\mathrm{m} \end{gathered}$$

Hence, the value of the horizontal component is $96.2\mathrm{m}$.

Hence vertical distance covered in 5.0s is given using equation (ii). Substitute the values in the equation (ii) to find the values of y-components as:

$$\begin{gathered} y=0.762\mathrm{m}+\left(27.47\mathrm{m}/\mathrm{s}脳5\mathrm{s}\right)+\left(\frac{1}{2}\right)脳\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)脳{\left(5\mathrm{s}\right)}^2 \\ =15.612\mathrm{m} \end{gathered}$$

But the vertical distance by which the ball clears the wall is given as:

$$\begin{gathered} d=15.612\mathrm{m}-11.28\mathrm{m} \\ =4.33\mathrm{m} \end{gathered}$$

Hence, the value of the vertical distance is $4.33\mathrm{m}$.

Now for time,

$$\begin{gathered} t\text{'}=5\mathrm{s}-0.5\mathrm{s} \\ =4.5\mathrm{s} \end{gathered}$$

The horizontal displacement is calculated by substituting the values of initial velocity and time in equation (ii).

$$\begin{gathered} 鈭�x\text{'}=V_{0x}脳t\text{'} \\ =19.23\mathrm{m}/\mathrm{s}脳4.5\mathrm{s} \\ =86.5\mathrm{m} \end{gathered}$$

The vertical displacement (y) is found using equation (ii). Substitute the values in the equation (ii).

$$\begin{gathered} y-0.762\mathrm{m}=\left(27.47\mathrm{m}/\mathrm{s}脳4.5\mathrm{s}\right)+\left(\frac{1}{2}\right)\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)脳{\left(4.5\mathrm{s}\right)}^2 \\ \mathrm{y}=25.152\mathrm{m} \\ 鈮�25.15\mathrm{m} \end{gathered}$$

Hence, the horizontal and vertical displacements of the ball with respect to home plate are 86.5 m, and 25.15 m respectively.