Q18E - University Physics with Modern Physics

Question

A shot putter releases the shot some distance above the level ground with a velocity of $\mathbf{12}\mathit{m}\mathbf{/}\mathit{s}\mathbf{,}\mathbf{51}\mathbf{°}$ above the horizontal. The shot hits the ground 2.08 s later. Ignore air resistance. (a) What are the components of the shot’s acceleration while in flight? (b) What are the components of the shot’s velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally? (d) Why does the expression for R in Example 3.8 not give the correct answer for part (c)? (e) How high was the shot above the ground when she released it? (f) Draw x-t, y-t, ${\mathit{v}}_{\mathbf{x}}\mathbf{-}\mathit{t}$ , and ${\mathit{v}}_{\mathbf{y}}\mathbf{-}\mathit{t}$ graphs for the motion.

Step-by-Step Solution

Answer

a) The horizontal and vertical component of the acceleration of shot is 0 and $-9.8\mathrm{m}/{\mathrm{s}}^2$ respectively.

b) The velocities in the beginning are 7.55 m/s and 9.3 m/s respectively, and the components of velocity at the end of trajectory are 7.33 m/s and -11.1 m/s respectively.

c) The distance of shot from the point it is thrown is 15.7 m .

d) The correct answer is given by $x=v_{x}t$ because the starting and finish point of the shot is different.

e) The height of shot above the ground when its thrown is 1.80 m .

f) The graphs for motion are given by,

Step 1: Identification of given data

The given data can be listed below,

• The velocity of the shot is, ${\mathrm{v}}_0=12 \mathrm{m}/\mathrm{s}$
• The angle of the shot above the horizon is $\theta =51°$ .
• The time at which shot hits the ground is $t=2.08 s$ .

Step 2: Concept/Significance of projectile motion.

An object moves along a predictable route in projectile motion, which is only affected by the initial launch speed, launch angle, and the acceleration from gravity.

Step 3: (a) Determination of the components of the shot’s acceleration while in flight

The component of acceleration in horizontal direction is zero due to the negligible air resistance whereas the vertical component of acceleration is only applicable in this condition in downward direction and its value is given by,

$$\begin{gathered} a_y=-g \\ =-9.8 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the horizontal and vertical component of the acceleration of shot is 0 and $-9.8 \mathrm{m}/{\mathrm{s}}^2$ respectively.

Step 4: (b) Determination of the components of the shot’s velocity at the beginning and at the end of its trajectory

The initial velocity of the shot in horizontal direction is given by,

$v_{0x}=v_0\cos \theta$

Here, $v_0$ is the velocity of the shot

Substitute all the values in the above,

$$\begin{gathered} v_{ox}=\left(12\mathrm{m}/\mathrm{s}\right)\cos 51° \\ =7.55 \mathrm{m}/\mathrm{s} \end{gathered}$$

The initial vertical component of the velocity of the shot is given by,

$v_{0y}=v_0\sin \theta$

Substitute all the values in the above,

$$\begin{gathered} v_{0y}=\left(12\mathrm{m}/\mathrm{s}\right)\sin 51° \\ =9.33\mathrm{m}/\mathrm{s} \end{gathered}$$ 

The final horizontal velocity of the ball at end of trajectory is given by,

$v_x=v_{0x}$

Substitute value in the above,

$v_x=7.55\mathrm{m}/\mathrm{s}$

The final vertical velocity of the ball at end of the trajectory is given by,

$v_y=v_{0y}-gt$

Here, $v_{0y}$ is the vertical component of initial velocity, t is the flight time, -g is the acceleration in vertical direction.

Substitute all the values in the above,

 $$\begin{gathered} v_y=9.33\mathrm{m}/\mathrm{s}-\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2.08\mathrm{s}\right) \\ =-11.1 \mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocities in the beginning are 7.55 m/s and 9.33 m/s respectively, and the components of velocity at the end of trajectory are 7.33 m/s and -11.1 m/s respectively.

Step 5: (c) Determination of the distance of the shot thrown horizontally

The horizontal distance of the shot is given by,

$x=v_{x}t$

Here, $v_x$ is the horizontal velocity of the shot and it is the time of flight.

Substitute all the values in the above expression.

$$\begin{gathered} x=\left(7.55m/s\right)\left(2.08s\right) \\ =15.7 m \end{gathered}$$

Thus, the distance of shot from the point it is thrown is 15.7 m .

Step 6: (d) Determination of the expression for R in Example 3.8 not give the correct answer for part (c)

The start point and the finish point are different because the shot was fired from a starting location that was elevated above the surface. So, the correct answer is given by equation $x=v_{x}t$ not by $x=\frac{v_0^2\sin 2\theta }{g}$ .

Thus, the correct answer is given by $x=v_{x}t$ because the starting and finish point of the shot is different.

Step 7: (e) Determination of the height of the shot above the ground when it released.

The equation for the displacement of the shot is given by,

$y=y_0+v_{0y}t-\frac{1}{2}g{t}^2$

Here, $y_0$ is the initial position of shot, $v_{oy}$ is the vertical component of initial velocity, t is the flight time, -g is the acceleration in vertical direction.

Substitute all the values in the above,

$$\begin{gathered} y_0=0-\left(12\sin 51°\right)\left(2.08\mathrm{s}\right)+\frac{1}{2}\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right){\left(2.08 \mathrm{m}/{\mathrm{s}}^2\right)}^2 \\ =-9.33 \mathrm{m}/\mathrm{s}\times 2.08 \mathrm{s}+4.9\mathrm{m}/{\mathrm{s}}^2{\left(2.08\mathrm{s}\right)}^2 \\ =1.80 \mathrm{m} \end{gathered}$$

Thus, the height of shot above the ground when its thrown is 1.80 m .

Step 8: (f) Sketch of the x-t, y-t, v x - t , and v y - t graphs for the motion.

The expression for horizontal displacement as a function of time is given by,

$$\begin{gathered} x=v_{x}t \\ =7.55 t \end{gathered}$$

The expression for vertical displacement as a function of time is given by,

$y=y_0+v_{0y}t-\frac{1}{2}g{t}^2$

Solve the above expression

 $y=1.80m+\left(9.33m/s\right)t-\left(4.9m/s^2\right)t^2$

The horizontal velocity of the shot is given by,

$$\begin{gathered} v_x=v_{0x} \\ =7.55m/s \end{gathered}$$

The vertical velocity of the shot is given by,

$$\begin{gathered} v_y=v_{oy}-gt \\ =9.33m-\left(9.8 m/s^2\right)t \end{gathered}$$

The graphs for the following are given by,