Question

a ball thrown verticaly upwards returnes to the theower after 10s find maximum height it reaches
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Answer 1

Explanation:

Hope it will help you!!!

Answer 2

Answer:

$the \: maximum \: height \: the \: ball \: \\ attains = 122.5m$

Explanation:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: given$

$total \: time \: of \: journey = 10sec$

$time \: taken \: to \: cover \: upward \: \\ journey = \frac{10}{2} = 5sec \\ (since \: time \: of \: ascent = time \: of \: \\ descent)$

$initial \: velocity(u) = ?$

$final \: velocity = 0$

$taking \: \: g \: (acceleration \: due \: to \: \\ gravity) = - 9.8ms {}^{ - 2}$

$using \: first \: equation \: of \: motion \\ = > v = u + gt$

$= > 0 = u + (- 9.8) \times 5$

$= > u = 49ms {}^{ - 1}$

$let \: the \: maximum \: height \\ \: secured \: by \: the \: ball \: when \: \\ thrown \:upwards \: be \: = h$

$using \: second \: equation \: of \\ \: motion = > h = ut + \frac{1}{2} gt {}^{2}$

$= > h = 49 \times 5 + \frac{1}{2} \times ( - 9.8) \times( 5 ){}^{2}$

$= > h = 245 - 122.5$

$= > h = 122.5m$

$thus \: the \: maximum \: height \: the \: \\ ball \: reaches = 122.5m$

hope this helps V●ᴥ●V