Question

A body is thrown vertically upwards and takes 5 seconds to reach maximum height. The distance travelled by the body will be same in
(1) 1st &10 th sec
(2) 2nd & 8 th sec
(3) 4th & 6th sec
(4) Both (2) &(3)

Answer 1

Answer on Question #45186, Physics, Other

A body is thrown vertically upwards and takes 5 seconds to reach maximum height. The distance travelled by the body will be same in

(1) 1st & 10th sec

(2) 2nd & 8th sec

(3) 4th & 6th sec

(4) Both (2) & (3)

Solution.

Let's calculate distances travelled by the body in n-th second.

$$S _ {n} = V _ {n - 1} \Delta t - \frac {g \Delta t ^ {2}}{2}$$

Let's find $V_0$ - the initial velocity of the body:

$$0 = V _ {0} - g \cdot 5 \Delta t \rightarrow V _ {0} = 5 g \Delta t$$

Where $V_{n-1}$ is a projection of body's velocity on the vertical axis upwards, $\Delta t = 1s$ .

$$V _ {n - 1} = V _ {0} - g t _ {n - 1} = V _ {0} - g (n - 1) \Delta t$$

Where (n-1) is a number of seconds passed before beginning of n-th second. So:

$$\begin{array}{l} S _ {n} = (V _ {0} - g (n - 1) \Delta t) \Delta t - \frac {g \Delta t ^ {2}}{2} = V _ {0} \Delta t - n g \Delta t ^ {2} + \frac {g \Delta t ^ {2}}{2} \\ = 5 g \Delta t ^ {2} - n g \Delta t ^ {2} + \frac {g \Delta t ^ {2}}{2} \\ \end{array}$$

Let $\Delta t$ be 1:

$$S _ {n} = 5 g - n g + \frac {g}{2} = \frac {1 1 g}{2} - n g$$

Here the dimension of $g$ is meters.

So distances traveled by the body are:

As we can see, from proposed answers only first is right: absolute amount of distances travelled in 1st &10 th seconds are equal.

Answer: (1) 1st &10 th sec

http://www.AssignmentExpert.com/