Question

50 points....


full explanation...


With what speed should a body be thrown upwards so that
the distances traversed in 5th second and 6th second are
equal?
(a) 5.84 ms-1 (b) 49 ms (0) 138 ms (0) 98 ms
​

Answer 1

SOLUTION:-

Given,

Distance travelled in the 5th second, PQ

$= > PQ = D5 = u + \frac{a}{2} (2n - 1) \ \\ = > PQ = u - \frac{g}{2} (2 \times 5 - 1).....(vertical \: upward \: motion \: of \: the \: body \: a = - g) \\ = > PQ = u - \frac{9g}{2} ...........(eq.1 )$

Now the distance QP is given as:

$QP = ut + \frac{1}{2} g {t}^{2} \\ = > QP = 0 + \frac{1}{2} g(1) {}^{2} \\ = > QP = \frac{g}{2} ..........(eq.2)$

Now, PQ= QP

So from equation (1) & (2) we get:

$= > u - \frac{9g}{2} = \frac{g}{2} \\ \\ = > u = 5g \\ \\ = > 5 \times 9.8 = > 49m/s \\ \\ = > u = 49 m/s$

Option (b)✓

Hope it helps ☺️

Answer 2

Refer to the attachment ❤️