Question

Suppose aball is thrownvertically upwards andreturns to the thrower after 12sec(g=9.8 m/s2)Find: (i) The velocity with which it was thrownup.[58.8m/s](ii) The maximum height it reaches.[176.4m](iii) Its position after 6secondand 9secon

Answer 1

Given:

Suppose a ball is thrown vertically upwards and returns to the thrower after 12sec(g=9.8 m/s²).

To find:

(i) The velocity with which it was thrownup.[58.8m/s]

(ii) The maximum height it reaches.[176.4m]

(iii) Its position after 6 secondand 9 seconds.

Calculation:

Since the time for total journey is 12 seconds , hence the time taken for upward journey will be 12/2 = 6 secs.

So, let initial velocity be u :

$\therefore \: v = u + at$

$= > \: 0 = u + ( - g)t$

$= > \:u = gt$

$= > \:u = 9.8 \times 6$

$\boxed{ = > \:u = 58.8 \: m {s}^{ - 1} }$

Let maximum height be h :

$\therefore \: h = ut + \dfrac{1}{2} a {t}^{2}$

$= > \: h = (58.8 \times 6)+ \dfrac{1}{2} ( - 9.8) {(6)}^{2}$

$= > \: h = 352.8 - 176.4$

$\boxed{ = > \: h = 176.4 \: m}$

So, after 6 seconds , object will be at its highest position i.e. at 176.4 m

After 9 secs , let height be H:

$\therefore \: H = ut + \dfrac{1}{2} a {t}^{2}$

$= > \: H = (58.8 \times 9) + \dfrac{1}{2} ( - 9.8){(9)}^{2}$

$= > \: H = 529.2 - 396.9$

$\boxed{ = > \: H = 132.3 \: m}$

HOPE IT HELPS.