Question

A physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in the figure. The ball's position is shown at 1s intervals until t=3s. At t=1s , the ball's velocity is v = (2.0 i + 2.0 j) m/s.

A. Determine the ball's velocity at t = 0s

B. Determine the ball's velocity at t = 2s

C. Determine the ball's velocity at t = 3s

D. What is the value of g on planet Exidor?

E. What was the ball's launch angle?

Thanks!

Answer 1

Concept and reason

The concept behind this question is Projectile motion and kinematic equations.

Find the velocities of projectile at the given points by using the kinematic equations.

Find the ball’s launch angle by using the relation among the tangent angle, vertical component of speed, and horizontal component of speed.

Fundamentals

An object or particle is thrown near Earth’s surface, and it moves along a curved path under the action of gravity is known as projectile motion.

The velocity vector of projectile can be expressed as follows:

$\overrightarrow v = {v_x}\hat i + {v_y}\hat j$

Here, $\hat i$ and $\hat j$ are the unit vectors in horizontal and vertical directions respectively, ${v_x}$ is the horizontal component, and ${v_y}$ is the vertical component.

The equation of projectile motion is,

${v_f} = {v_i} + at$

Here, ${v_i}$ is the initial velocity, ${v_f}$ final velocity, a acceleration due to gravity, and t is time.

The launch angle is expressed as,

$\theta = {\tan ^{ - 1}}\left( {\frac{{{v_y}}}{{{v_x}}}} \right)$

Here, ${v_y}$ is the y-component of given velocity vector and ${v_x}$ is the x-component of the given velocity vector.

(A)

Determine the gravity of planet as follows:

$g = \frac{{{v_{2y}} - {v_{1y}}}}{{{t_2} - {t_1}}}$

Here, ${v_{1y}}$ and ${v_{2y}}$ are the vertical component of velocities at points 1 and 2 respectively, ${t_1}$ and ${t_2}$ are the corresponding times.

Consider the particle is moving from point 1 to point 2. The ball reached maximum height at point 2. At the maximum height, the vertical component of velocity becomes zero.

Substitute 0 for ${v_2}$ , $2.0{\rm{ m/s}}$ for ${v_{1y}}$ , 2 s for ${t_2}$ , and 1 s for ${t_1}$ in the expression $g = \frac{{{v_{2y}} - {v_{1y}}}}{{{t_2} - {t_1}}}$ to find gravity.

$\begin{array}{l}\\g = \frac{{0 - 2.0{\rm{ m/s}}}}{{2{\rm{ s}} - 1{\rm{ s}}}}\\\\{\rm{ }} = - 2.0{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The horizontal component of velocity remains same at any point along the path followed by the projectile. So,

${v_{1x}} = {v_{0x}}$

The velocity at 1 s is given as follows:

$\vec v = \left( {2.0\hat i + 2.0\hat j} \right){\rm{ m/s}}$

From the above equation, the x-component of velocity at 1 s is,

${v_{1x}} = 2.0{\rm{ m/s}}$

Substitute $2.0{\rm{ m/s}}$ for ${v_{1x}}$ in ${v_{1x}} = {v_{0x}}$ .

$\begin{array}{c}\\2.0{\rm{ m/s}} = {v_{0x}}\\\\{v_{0x}} = 2.0{\rm{ m/s}}\\\end{array}$

Consider the particle is moving from point 0 to point 1. Let at point 0 s vertical velocity component be ${v_{0y}}$ .

The vertical component of velocity at point 1s can be expressed in terms of vertical component of velocity at point 0s, acceleration a, and time t as follows:

${v_{1y}} = {v_{0y}} + at$

Substitute $2{\rm{ m/s}}$ for ${v_{0y}}$ , -2 for a, and 1s for t in ${v_{1y}} = {v_{0y}} + at$ to find the initial vertical velocity ${v_{0y}}$ .

$\begin{array}{c}\\2{\rm{ m/s}} = {v_{0y}} + \left( { - 2{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {1{\rm{ s}}} \right)\\\\{v_{0y}} = 4{\rm{ m/s}}\\\end{array}$

The velocity vector at 0 s can be expressed as follows:

${\vec v_0} = {v_{0x}}\hat i + {v_{0y}}\hat j$

Substitute $2.0{\rm{ m/s}}$ for ${v_{0x}}$ and $4{\rm{ m/s}}$ for ${v_{0y}}$ .

$\begin{array}{c}\\{{\vec v}_0} = \left( {2.0{\rm{ m/s}}} \right)\hat i + \left( {4{\rm{ m/s}}} \right)\hat j\\\\ = \left( {2.0\hat i + 4.0\hat j} \right){\rm{ m/s}}\\\end{array}$

(B)

At time 2 s, the object has reached the maximum height. The vertical component of velocity ${v_{2y}}$ at the maximum height is equal to zero.

${v_{2y}} = 0$

But, however the horizontal component of velocity remains constant at any point along the path followed by the projectile.

Thus, at the maximum height position, the velocity vector consists only of horizontal component.

Therefore, the velocity of the projectile at 2 s position is,

$\overrightarrow {{v_2}} = \left( {2.0\hat i} \right){\rm{ m/s}}$

(C)

Determine the velocity of the ball at t = 3s as follows:

Consider the particle is moving from point 0 to point 3.

Consider at point 0 s vertical velocity component is,

${v_{0y}} = 4.0{\rm{ m/s}}$

Let the vertical velocity component at point 3 s be ${v_{3y}}$ .

The vertical component of velocity at point 3 s is,

${v_{3y}} = {v_{0y}} + at$

The velocity at point 3 s can be expressed as follows:

${\vec v_3} = {v_{3x}}\hat i + {v_{3y}}\hat j$

Substitute $4.0{\rm{ m/s}}$ for ${v_{0y}}$ , $- 2.0{\rm{ m/}}{{\rm{s}}^2}$ for a, and 3s for t in ${v_f} = {v_i} + at$ to find the initial vertical velocity.

$\begin{array}{l}\\{v_{3y}} = 4.0{\rm{ m/s}} + \left( { - 2.0{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {3{\rm{ s}}} \right)\\\\{v_{3y}} = - 2.0{\rm{ m/s}}\\\end{array}$

The horizontal component of velocity at point 3 s is,

${v_{3x}} = 2.0{\rm{ m/s}}$

Substitute $2.0{\rm{ m/s}}$ for ${v_{3x}}$ and $- 2.0{\rm{ m/s}}$ for ${v_{3y}}$ in ${\vec v_3} = {v_{3x}}\hat i + {v_{3y}}\hat j$ .

$\begin{array}{c}\\{{\vec v}_3} = \left( {2.0{\rm{ m/s}}} \right)\hat i + \left( { - 2.0{\rm{ m/s}}} \right)\hat j\\\\ = \left( {2.0\hat i - 2.0\hat j} \right){\rm{ m/s}}\\\end{array}$

(D)

Determine the gravity of planet as follows:

$g = \frac{{{v_{2y}} - {v_{1y}}}}{{{t_2} - {t_1}}}$

Here, ${v_{1y}}$ and ${v_{2y}}$ are the vertical component of velocities at points 1 and 2 respectively, ${t_1}$ and ${t_2}$ are the corresponding times.

Consider the particle is moving from point 1 to point 2. The ball reached maximum height at point 2. At the maximum height, the vertical component of velocity becomes zero.

Substitute 0 for ${v_2}$ , $2.0{\rm{ m/s}}$ for ${v_{1y}}$ , 2 s for ${t_2}$ , and 1 s for ${t_1}$ in the expression $g = \frac{{{v_{2y}} - {v_{1y}}}}{{{t_2} - {t_1}}}$ to find gravity.

$\begin{array}{l}\\g = \frac{{0 - 2.0{\rm{ m/s}}}}{{2{\rm{ s}} - 1{\rm{ s}}}}\\\\{\rm{ }} = - 2.0{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Therefore, the value of g is $- 2.0{\rm{ m/}}{{\rm{s}}^2}$ .

(E)

Determine the ball’s launch angle as follows:

Initial velocity of the ball is,

$\overrightarrow {{v_0}} = \left( {2.0\hat i + 4.0\hat j} \right){\rm{ m/s}}$

Substitute $2.0{\rm{ m/s}}$ for ${v_x}$ and $4.0{\rm{ m/s}}$ for ${v_y}$ in the equation

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{4.0{\rm{ m/s}}}}{{2.0{\rm{ m/s}}}}} \right)\\\\ = 63.43^\circ \\\end{array}$

Therefore, ball’s launch angle is $63.43^\circ$ .

Ans: Part A

The ball’s velocity at time t = 0s is $\left( {2.0\hat i + 4.0\hat j} \right){\rm{ m/s}}$ .

Part B

The ball’s velocity at time t = 2s is $\left( {2.0\hat i} \right){\rm{ m/s}}$

Part C

The ball’s velocity at time t = 3s is $\left( {2.0\hat i - 2.0\hat j} \right){\rm{ m/s}}$

Part D

The value of g on planet Exidor is $- 2.0{\rm{ m/}}{{\rm{s}}^2}$ .

Part E

The ball’s launch angle is $63.43^\circ$