Question

A farm truck moves due east with a constant velocity of $\mathbf{9}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$on a limitless, horizontal stretch of road. A boy riding on the back of the truck throws a can of soda upward (Fig. P4.54) upward (Fig. P4.54) truck bed, but $\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$farther down the road. (a) In the frame of reference of the truck, at what angle to the vertical does the boy throw the can? (b) What is the initial speed of the can relative to the truck? (c) What is the shape of the can's trajectory as seen by the boy? An observer on the ground watches the boy throw the can and catch it. In this observer's frame of reference, (d) describe the shape of the can's path and (e) determine the initial velocity of the can.

Short answer

(a) The vertical angle with which the boy throws the can in the frame of reference of the truck is $0°$.

(b) The initial velocity of the can of soda relative to the truck is $8.23\text{m}/\text{s}$.

(c) The shape of the can of soda's trajectory as seen by the boy is a straight line.

(d) In the ground frame of reference shape of Can’s path is Parabolic.

(e) The initial velocity of the can is $12.6 \mathrm{m}/\mathrm{s}$.

Step-by-step solution

Understanding the Concept

The expression for the horizontal motion from the Newton’s second law is given by,

$x=x_i+v_{i}t+\frac{1}{2}a{t}^2$ ...... (1)

Here,$x$is the distance,$x_i$is the initial distance,$v_i$is the velocity of the truck, a is the acceleration, t is the time taken.

The path followed by the moving object is known as the trajectory.

(a) Vertical angle 

The velocity of truck is $9.50m/s$and the distance is 16 m.

If the can is thrown at an angle, it will have both horizontal and vertical components of the velocity.

A nonzero horizontal component would lead the ball to move an extra distance than due to the motion of the car. Hence, it cannot be caught at the same location on the truck bed.

Thus, the can of soda must be thrown straight up at $0°$to the vertical for the boy to catch it at the same location on the truck bed.

Therefore, the vertical angle with which the boy throws the can of soda in the frame of reference of the truck is $0°$.

(b) Initial speed of the can relative to the truck

Substitute 16 m for x, 0 for $x_i$for and 0 for $v_i$ in equation (1) to find t.

$\begin{array}{rcl}16\text{m}& =& 0+(9.50\text{m}/\text{s})t+\frac{1}{2}\left(0\right)t^{2}16\text{m}\\ & =& (9.50\text{m}/\text{s})t\end{array}$

Solve for t.

$\begin{array}{rcl}t& =& \frac{16\text{m}}{9.50\text{m}/\text{s}}\\ & =& 1.68\text{s}\end{array}$

Write the equation for the free fall of can of soda is written

$y_{\text{f}}=y_{\text{i}}+v_{\text{yi}}t+\frac{1}{2}a{t}^2$

Here, is the initial velocity of the can of soda.

$v_{yi}$

Substitute 0 for $y_{\text{f}}$and $y_{\text{i}},1.68\text{s}$ for t and $-9.8\text{m}/{\text{s}}^2$ for a in above equation to find $v_{\text{yi}}$,

$\begin{array}{rcl}0& =& 0+v_{\text{yi}}(1.68\text{s})+\frac{1}{2}\left(-9.8\text{m}/{\text{s}}^2\right)(1.68\text{s}{)}^2{v}_{\text{yi}}(1.68\text{s})\\ v_{\text{yi}}& =& 8.23\text{m}/\text{s}\end{array}$

Therefore, the initial velocity of the can of soda relative to the truck is $8.23\text{m}/\text{s}$.

(c) Shape of the Can’s trajectory as seen by the boys

The boy sees the can of soda always over his head, traversing a straight-line segment upward and the downward.

Thus, its trajectory is a straight line.

Therefore, the shape of the can of soda's trajectory as seen by the boy is a straight line.

(d) Shape of the Can’s path

In the of the can velocity in ground frame of reference, the horizontal velocity of the can of soda is equal to the velocity of the truck.

$v_x=9.50 m/s$

In the ground frame of reference shape of Can’s path is Parabolic.

(e) Shape of the Can’s path

In the ground frame of reference, the can has both horizontal and vertical component of velocity. The result value of initial velocity is given by,

$\begin{array}{rcl}{v}_i& =& \sqrt{v_x^2+v_y^2}\\ & =& \sqrt{9.{50}^2+8.{25}^2}\\ & =& 12.6 m/s\end{array}$

Therefore the initial velocity of the can is $12.6 m/s$.