Question

a stone is projected vertically upward if it travels equal distance in fourth second and fifth second then find the time of flight of stone​

Answer 1

The body of thrown vertically upward. This means that it goes under constant acceleration of 9.8 m/sec²

To make calculations comfortable and easy, we will assume g as 10 m/sec².

Now,

For a constantly accelerating body, distance in two consecutive seconds can only remain same when body retraces its path travelled in 1st second in its 2nd second.

$\rule{200}{2}$

From the figure, the calculations proceed as follows :-

We know :-

$s_{n\:th}= u + \dfrac{a}{2}(2n-1)$

Now

$u + \frac{10}{2} \times 7 = 0 + \frac{10}{2} \times 9 \\ \\ \\u = 45 - 35 \\ \\ \\u = 10$

Hence,

Initial velocity is 10 m/sec

Now,

Final velocity will always be equal to 0 m/sec.

Now,

Total distance it will travel will be :-

$= 10 \times 4 + \frac{1}{2} \times 10 \times 16 \\ \\ \\ = 40 + 80 \\ \\ \\ = 120m$

Now,

Getting time to reach top :-

$s = ut + \frac{1}{2}a {t}^{2} \\ \\120 = 10t + 5 {t}^{2} \\ \\ \\ {t}^{2} + 2t - 24 = 0 \\ \\ \\t = 4 \\ \\t = ( - 6) \: \: \: \: \: neglected$

Thus,

Time to reach top = 4 seconds.

Total flight of stone = 2 * time to reach top

So,

Time will be

$= 2 \times 4 \\ \\ = 8sec$

$\rule{200}{2}$

Answer 2

Answer:-

u+ 210 ×7=0+ 210×9

u=45−35

u=10

Velocity =10 m/sec

Velocity equals --m/sec.

Distance travel

=10×4+ 21 ×10×16

=40+80

=120m

s=ut+ 212

120=10t+5t

t 2 +2t−24=0

t=4

t=(−6)

Time = 4 seconds.

Flight ✈ = 2 to

=2×4

=8sec