Question

A home run is hit in such a way that the baseball just clears a wall 21 m tall, located at a distance 130 m from the home plate. the ball is hit at a angle of 35 deg to the horizontal and there is no air resistance. Find(a) the initial speed of the ball (b) the time it takes to reach the wall and (c) the velocity components and the speed of the ball when it reaches the wall.( assume the ball was hit at a height of 1 m above the ground. Please show me the steps you take to solve this. Thank you

Answer 1

Concepts and reason

The concepts used to solve the question are the equations of the motion and the projectile motion.

At first determine the time taken by the ball to reach the fence. Then use this time to determine the initial velocity of the ball at which it is being launched. Finally determine the velocity of the ball when it reaches to the wall.

Fundamentals

Equations of the motion

An object is said to be in motion when it changes its position. This change in position of the object is describe in terms of displacement, speed, time and acceleration of the object.

The equations describing the position and the velocity as a function of time are termed as equations of the motion. An object moving under constant acceleration is described under these set of equations.

The equations of the motion are defined as,

$v = u + at$

Here, $v$ is the final velocity of the particle, $u$ is the initial velocity, $a$ is the uniform acceleration of the particle and $t$ is the time taken.

$S = ut + \frac{{a{t^2}}}{2}$

Here, $S$ is the displacement of the particle.

${v^2} - {u^2} = 2aS$

In projectile motion the only force acting on the object is the force of gravity. Hence, the acceleration of the object in the vertical direction is the acceleration due to gravity.

${a_y} = \pm g$

Here, the acceleration due to gravity is positive if the motion of the object is downwards and negative if the motion of the object is upwards.

(a)

The time taken by the ball to reach the fence is,

$t = \frac{{\Delta x}}{{{v_{{\rm{ix}}}}}}$ …… (1)

Here, $\Delta x$ is the horizontal distance travelled by the ball and ${v_{{\rm{ix}}}}$ initial horizontal component of the velocity.

The height of the ball from the ground where it is being launched is

${y_0} = 1{\rm{ m}}$

The initial horizontal component of the velocity of the ball is,

${v_{{\rm{ix}}}} = {v_{\rm{i}}}\cos \theta$

The initial vertical component of the velocity of the ball is,

${v_{{\rm{iy}}}} = {v_{\rm{i}}}{\rm{s}}in\theta$

Here, ${v_i}$ is the initial velocity of the ball and $\theta$ is the angle of projection.

From the equation of motion.

$y - {y_0} = {v_i}t + \frac{{{a_y}{t^2}}}{2}$ …… (2)

Substitute, equation (1) in (2).

$y - {y_0} = {v_{iy}}t + \frac{{{a_y}{t^2}}}{2}$

$y - {y_0} = {v_{iy}}\left( {\frac{{\Delta x}}{{{v_{ix}}}}} \right) + \frac{{{a_y}{{\left( {\frac{{\Delta x}}{{{v_{ix}}}}} \right)}^2}}}{2}$ …… (3)

Substitute, the values of ${v_{{\rm{ix}}}}{\rm{ and }}{v_{{\rm{iy}}}}$ in the equation (3).

$\begin{array}{c}\\y - {y_0} = {v_{iy}}\left( {\frac{{\Delta x}}{{{v_{ix}}}}} \right) + \frac{{{a_y}{{\left( {\frac{{\Delta x}}{{{v_{ix}}}}} \right)}^2}}}{2}\\\\ = \left( {{v_i}\sin \theta } \right)\left( {\frac{{\Delta x}}{{{v_i}\cos \theta }}} \right) + \frac{{{a_y}{{\left( {\frac{{\Delta x}}{{{v_i}\cos \theta }}} \right)}^2}}}{2}\\\end{array}$

$y - {y_0} = \Delta x\tan \theta + \frac{{{a_y}\Delta {x^2}}}{{2{{\left( {{v_i}\cos \theta } \right)}^2}}}$ …… (4)

Substitute $130{\rm{ m}}$ for $\Delta x$ , $1{\rm{ m}}$ for ${y_0}$ , $21{\rm{ m}}$ for $y$ $35^\circ$ for $\theta$ and $- 9.81{\rm{ m/}}{{\rm{s}}^2}$ for ${a_y}$ in the equation (4).

$\begin{array}{c}\\y - {y_0} = \Delta x\tan \theta + \frac{{{a_y}\Delta {x^2}}}{{2{{\left( {{v_i}\cos \theta } \right)}^2}}}\\\\21{\rm{ m}} - 1{\rm{ m}} = \left( {130{\rm{ m}}} \right)\tan 35^\circ - \frac{{\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right){{\left( {130{\rm{ m}}} \right)}^2}}}{{2{{\left( {{v_i}\cos 35^\circ } \right)}^2}}}\\\\20{\rm{ m }} = 91.026{\rm{ m }} - \frac{{123538.7481{\rm{ }}{{\rm{m}}^3}/{{\rm{s}}^2}}}{{{v_i}^2}}\\\\{v_i}^2 = \frac{{123538.7481{\rm{ }}{{\rm{m}}^3}/{{\rm{s}}^2}}}{{71.026{\rm{ m}}}}\\\end{array}$

Therefore,

$\begin{array}{c}\\{v_i}^2 = \frac{{123538.7481{\rm{ }}{{\rm{m}}^3}/{{\rm{s}}^2}}}{{71.026{\rm{ m}}}}\\\\ = \sqrt {1739.345{\rm{ }}{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}} \\\\ = 41.705{\rm{ m/s}}\\\end{array}$

(b)

The taken by the ball to cover the horizontal distance of $130{\rm{ m}}$ is,

$t = \frac{{\Delta x}}{{{v_{\rm{i}}}\cos \theta }}$ …… (5)

Substitute, $130{\rm{ m}}$ for $\Delta x$ , $35^\circ$ for $\theta$ and $41.705{\rm{ m/s}}$ for ${v_{\rm{i}}}$ in the equation (5).

$\begin{array}{c}\\t = \frac{{\Delta x}}{{{v_{\rm{i}}}\cos \theta }}\\\\ = \frac{{130{\rm{ m}}}}{{\left( {41.705{\rm{ m/s}}} \right)\left( {{\rm{cos 35}}^\circ } \right)}}\\\\ = 3.80{\rm{ s}}\\\end{array}$

(c)

The final component of the velocity of the ball in the horizontal direction is,

${v_{{\rm{fx}}}} = {v_{\rm{i}}}\cos \theta$ …… (6)

Substitute $41.705{\rm{ m/s}}$ for ${v_{\rm{i}}}$ and $35^\circ$ for $\theta$ in the equation (6)

$\begin{array}{c}\\{v_{{\rm{fx}}}} = {v_{\rm{i}}}\cos \theta \\\\ = \left( {41.705{\rm{ m/s}}} \right)\cos 35^\circ \\\\ = 34.162{\rm{ m/s}}\\\end{array}$

The final vertical component of the velocity is,

${v_{{\rm{fy}}}} = {v_{\rm{i}}}\sin \theta - gt$ …… (7)

Substitute $41.705{\rm{ m/s}}$ for ${v_{\rm{i}}}$ , $35^\circ$ for $\theta$ , $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ and $3.80{\rm{ s}}$ for $t$ in the equation (7).

$\begin{array}{c}\\{v_{{\rm{fy}}}} = {v_{\rm{i}}}\sin \theta - gt\\\\ = \left( {41.705{\rm{ m/s}}} \right)\sin 35^\circ - \left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {3.80{\rm{ s}}} \right)\\\\ = - 13.356{\rm{ m/s}}\\\end{array}$

The resultant velocity of the ball when it reaches at the wall is,

$\begin{array}{c}\\v = \sqrt {{v_{{\rm{fx}}}}^2 + {v_{{\rm{fy}}}}^2} \\\\ = \sqrt {{{\left( {34.162{\rm{ m/s}}} \right)}^2} + {{\left( { - 13.356{\rm{ m/s}}} \right)}^2}} \\\\ = 36.680{\rm{ m/s}}\\\end{array}$

Ans: Part a

The initial speed of the ball is $41.705{\rm{ m/s}}$

Part b

The time taken by the ball to reach the wall is $3.80{\rm{ s}}$

Part c

The final velocity of the ball when it reaches at the wall is $36.680{\rm{ m/s}}$