Question

A science student is riding on a flatcar of a train traveling along a straight, horizontal track at a constant speed of 10 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 60.0⁰with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does she see the ball rise?

Short answer

The height of the ball is 15.3 m.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The speed of train is 10 m/s .
• The initial angle of ball with horizontal is60.0⁰

The significance of the relative velocity

The relative velocity is the velocity of an object or observer B in the rest frame of another object or observer A .

The expression for the  x component of velocity

Write the expression for the component of velocity

$v_{\text{ix}}=v_{\text{i}}\cos 60°$

Here, $v_{\text{i}}$is the initial speed of the ball with respect to the student and $v_{\text{ix}}$is the corresponding -component of the velocity

The y component of velocity

Write the expression for the y component of velocity

$v_{\text{iy}}=v_{\text{i}}\sin 60°$

Here, $v_{\text{iy}}$ is the y - component of the velocity of the ball with respect to the student

The relative velocity in the x-direction

Write the equation for relative velocity in the x-direction

$v_{\text{x}}=v_{\text{ix}}+v$

Here, $v_{\text{x}}$is the velocity of ball relative to the professor in the x-direction and v is the velocity of the student with respect to the professor.

There is no x motion reference to professor,

localid="1663857426127" $$\begin{gathered} v_{\text{x}}=v_{\text{ix}}+v=0 \\ {v}_{\text{ix}}=-v \end{gathered}$$

Substitute values in the above equation.

$v_{\text{ix}}=-(10 \text{m/s})$

ratio of  x component of velocity and y component of velocity

Write the expression for the ratio of x component of velocity and y component of velocity to find

$\begin{array}{rcl}\frac{v_{iy}}{v_{ix}}& =& \frac{v_i\sin 60°}{v_i\cos 60°}\\ & =& \tan 60°\\ & =& 1.73v_{iy}\\ & =& (1.73)v_{ix}\\ & & \end{array}$

The y-component of the velocity of the ball with respect to the student is same as the velocity of the ball as seen by the professor.

The height of ball

Write the formula to calculate height of ball

$h=\frac{v_{yi}^2}{2g}$

Substitute values in the above equation.

$h=\frac{{\left(1.73v_{ix}\right)}^2}{2g}$

Substitute values in the above equation to find h.

$\begin{array}{rcl}h& =& \frac{{\left(\right(1.73)\times (-10 \text{m/s}\left)\right)}^2}{2\left(9.8 {\text{m/s}}^2\right)}\\ h& =& 15.26 \text{m}\\ & \approx & 15.3 \text{m}\\ & & \end{array}$

Therefore, the height of the ball is 15.3 m .