Q12E - University Physics with Modern Physics

Question

A rookie quarterback throws a football with an initial upward velocity component of 12.0 m/s and a horizontal velocity component of 20 m/s. Ignore air resistance. (a) How much time is required for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after it is thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? (d) How far has the football traveled horizontally during this time? (e) 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 time is required for the football to reach the highest point of the trajectory is 1.22 s

b) The maximum height of the point is 7.34 m .

c) The time taken by the football to reach from maximum height to ground is 2.44 s.

d) The horizontal distance travelled by football is 49 m.

e) The graphs for x-t, y-t, $v_x-t$, $v_y-t$ of motion are

Step 1: Identification of given data

The given data can be listed below,

• The initial upward velocity is,$v_{0y}=12 \mathrm{m}/\mathrm{s}$
• The horizontal velocity component is,$v_{0x}=20 \mathrm{m}/\mathrm{s}$

Step 2: Concept/Significance of constant velocity.

When a moving body travels with constant velocity, it signifies that its speed is constant, non-varying, and constant with regard to time.

Step 3: Determination of time is required for the football to reach the highest point of the trajectory

(a)

 

The trajectory of football is shown below in the diagram,

                                         

The relation between initial and final velocity of the particle in vertical motion is given by,

$v_y=v_{0y}-gt$

Here, $v_y$ is vertical velocity at maximum height, $v_{0y}$is the vertical component of initial velocity, g is the acceleration due to gravity, and t is the time interval.

 

Substitute 12m/s for${\mathrm{v}}_{0\mathrm{y}}$and 9.8m/s² for g in the above equation.

$$\begin{gathered} 0=12\mathrm{m}/\mathrm{s}-\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)t \\ t=\frac{12\mathrm{m}/\mathrm{s}}{9.8 \mathrm{m}/{\mathrm{s}}^2} \\ =1.22 \mathrm{s} \end{gathered}$$

 

Thus, the time is required for the football to reach the highest point of the trajectory is 1.22 s

Step 4: Determination of the height of the point.

(b)

 

Equation of motion for constant acceleration is given by,

$v_y^2=v_{0y}^2-2gH$ 

Here, $v_{0y}$ is the vertical velocity at maximum height, $v_{0y}$ is the vertical component of initial velocity, g is the acceleration due to gravity, and H is the maximum height.

 

Substitute 12m/s for $v_{0y}$and $9.8\mathrm{m}/{\mathrm{s}}^2$ for $g$ in the above equation.

$$\begin{gathered} H=\frac{v_{0y}^2}{2g} \\ =\frac{{\left(12 \mathrm{m}/\mathrm{s}\right)}^2}{2\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)} \\ =7.34 \mathrm{m} \end{gathered}$$ 

 

Thus, the maximum height of the point is 7.34 m.

Step 5: Determination of time is required for the football to return to its original level.

(c)

 

The time taken by the football to reach from maximum height to ground is given by,

$T=2t$

Here, t is the time taken by football to reach maximum height.

 

Substitute $1.22s$ for t in the above equation.

$$\begin{gathered} T=2\left(1.22 s\right) \\ =2.44 s \end{gathered}$$

 

Thus, the time taken by the football to reach from maximum height to ground is 2.44 s .

Step 6: Determination of the distance of football traveled horizontally during the time.

(d)

 

The horizontal distance traveled by the football is given by,

$x=v_{0x}T$ 

Here, $v_{0x}$ is the horizontal component of velocity and T is the time taken by football to reach ground.

                             

Substitute 20m/s for $v_{0x}$ and 2.44s for $T$ in the above equation.

$$\begin{gathered} x=\left(20 \mathrm{m}/\mathrm{s}\right)\left(2.44 \mathrm{s}\right) \\ =49 \mathrm{m} \end{gathered}$$

 

Thus, the horizontal distance travelled by football is 49 m.

Step 7: Sketch of x -t, y -t, v x - t , v y - t and graphs for the motion.

(e)

 

The graphs for x-t, y-t, $v_x-t$, $v_y-t$ of motion are given below as,