Question

with what speed should be a body be thrown upwards so that the distance traversed in 5th sec and 6th sec are equal?

Answer 1

We will use the following equation of motion :

S = ut - 1 / 2gt²

Let g = 10m/s²

Since it is thrown up.

At the 5th second S will be :

S = 5u - 0.5 × 10 × 5²

S= 5u — 125

At the 6th second S will be :

S = 6u - 0.5 × 10 × 6²

S= 6u - 180

Since the two displacements are equal, we have :

6u - 180 = 5u - 125

u = 180 - 125 = 55m / s

Answer 2

"The body should be thrown at a speed of $53.9 \mathrm{ms}^{-1}$

Given:

Let the distance traversed in 5th second and 6th second be$\mathrm{s}_{5} and \mathrm{s}_{6}$ respectively.

$\mathrm{s}_{5}=\mathrm{s}_{6}$

Initial velocity, u = ?

Acceleration due to gravity, $g=9.8 \mathrm{ms}^{-2}$

Solution:

The distance travelled by a body thrown vertically upwards is

$s=u t-\left(\frac{g t^{2}}{2}\right)$

On substituting, we get

At 5th second,

$S_{5}=u \times 5-\left(\frac{9.8 \times 5^{2}}{2}\right)=5 u-122.5$

At 6th second,

$s_{6}=u \times 6-\left(\frac{9.8 \times 6^{2}}{2}\right)=6 u-176.4$

Since,$\mathrm{s}_{5}=\mathrm{s}_{6}$

$\Rightarrow 5 u-122.5=6 u-176.4$

On rearranging, we get,

6u-5u = 176.4-122.5

$\mathrm{u}=53.9 \mathrm{ms}^{-1}$"