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Q2.6(d):
A player throws a ball upwards with an initial speed of 29.4 m s(^{-1}). (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take (g) = 9.8 m s(^{-2}) and neglect air resistance).

Solution :

Step 1: Identifying Known Variables and Equations of Motion

To analyze the vertical motion of the ball, we define the following variables based on the given data and standard kinematic conventions ($g$ is negative because it acts against the upward motion):

  • Initial velocity ($u$): $29.4 \, \text{m s}^{-1}$
  • Final velocity at maximum height ($v$): $0 \, \text{m s}^{-1}$
  • Acceleration due to gravity ($g$): $-9.8 \, \text{m s}^{-2}$
  • Displacement ($s$): $h$ (maximum height)

We utilize the third equation of kinematics: $v^2 = u^2 + 2as$ [Per Kinematic Laws of Uniformly Accelerated Motion].

Step 2: Calculating Maximum Height ($h$)

Substituting the values into the equation:

$0^2 = (29.4)^2 + 2(-9.8)(h)$

$0 = 864.36 - 19.6h$

$19.6h = 864.36$

$h = \frac{864.36}{19.6} = 44.1 \, \text{m}$

Step 3: Calculating Total Time of Flight ($t$)

The time taken to reach the maximum height ($t_{up}$) can be determined using the first equation of motion: $v = u + at$.

$0 = 29.4 + (-9.8)(t_{up})$

$9.8t_{up} = 29.4$

$t_{up} = \frac{29.4}{9.8} = 3 \, \text{s}$

Since air resistance is neglected, the time taken to rise to the maximum height is equal to the time taken to fall back to the point of projection [By the Principle of Symmetry in Kinematics].

Total time of flight ($T$) = $t_{up} + t_{down} = 3 \, \text{s} + 3 \, \text{s} = 6 \, \text{s}$.

Visual Representation of Vertical Trajectory

h = 44.1 m Start/End Trajectory

Data Summary Table

Parameter Value
Maximum Height ($h$) $44.1 \, \text{m}$
Time to Max Height ($t_{up}$) $3.0 \, \text{s}$
Total Time of Flight ($T$) $6.0 \, \text{s}$

Final Solution: The ball rises to a maximum height of $44.1 \, \text{m}$ and returns to the player's hands after $6 \, \text{s}$.


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