Question

A cricket match customarily begins with a coin toss. The umpire tosses the coin up with an initial speed of $5.0 m s^{- 1}$ . In the absence of air resistance,

(a) how high does the coin go above its point of release?
(b) what is the total time the coin is in the air before returning to its release point?

Accepted Answer

(a) The statement, how high does the coin go, means the maximum height which occurs when the final velocity of the coin in the vertical direction is zero or, $v = 0 m s^{- 1}$ . Taking upward direction as positive, as the coin is tossed upward, hence, the initial velocity, $v_{0} = + 5.0 m s^{- 1}$ . The velocity of the coin is momentarily zero when the coin reaches its maximum height. There are three parts to the motion of the coin in which air resistance is being ignored.

(i) On the way up, the coin has an upward-pointing velocity vector with a decreasing magnitude.

(ii) At the top of its path, the velocity vector of the coin is momentarily zero.

(iii) On the way down, the coin has a downward-pointing velocity vector with an increasing magnitude.

In any of the cases, the object always experiences the same downward acceleration due to gravity. During the upward and downward parts of the motion and also at the top of the path, the acceleration due to gravity has a constant downward direction and a constant magnitude of $10 m s^{- 2}$ . In other words, the acceleration vector of the particle does not behave as the velocity vector does. In particular, the acceleration vector is not zero at the top of the motional path just because the velocity vector is zero there. Acceleration is the rate at which the velocity is changing and the velocity is changing at the top even though at one instant it is zero.

Using, $v^{2} = v_{0}^{2} + 2 a . s$ ,

$\Rightarrow 0 = (+ 5)^{2} + 2 (- 10) \cdot y$ which gives, $y = 1.25 m$ .

(b) During the time the coin travels upward, gravity causes its speed to decrease to zero. On the way down, however, gravity causes the coin to regain the lost speed. Thus, the time for the coin to go up is equal to the time for it to come down. In other words, the total travel time is twice the time for the upward motion.We can use equation, $v = v_{0} + a t$ to find the upward travel time,

$\Rightarrow t = \frac{v - v_{0}}{a} = \frac{0 - 5}{- 10} = 0.5 s$ .

The total up-and-down time is twice this value or $1.0 s$ .

It is possible to determine the total time by another method. When the coin is tossed upward and returns to its release point, the displacement for the entire trip is, $y = 0 m$ . With this value for displacement, we can use equation, $(y = v_{0} . t + \frac{1}{2} a . t^{2})$ to find the time for the entire trip directly,

$0 = (+ 5) \cdot T + \frac{1}{2} (- 10) \cdot T^{2} \Rightarrow T = 1 s$ .

B M Sharma Solutions for Chapter: Kinematics I, Exercise 26: ILLUSTRATION 4.24

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 26: ILLUSTRATION 4.24

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 26: ILLUSTRATION 4.24 with Hints & Solutions

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