Question

Example 6.13 A ball is thrown upwards from the top of a tower 40 m high
with a velocity of 10 m/s. Find the time when it strikes the ground.
Take g = 10 m/s.​

Answer 1

Answer:

$The \: ball \: will \: take \: 2 \: seconds \: to \: \\ strike \: the \: ground.$

Step-by-step explanation:

$Given, \\ Initial \: velocity \: (u) = 10 \: ms {}^{ - 1} \\ Acceleration \: ( g) = 10 \: ms{}^{ - 2} \\ Height \: of \: the \: tower \: (h) = 40 \: m \\ Using \: second \: equation \: of \: motion: \\ = > h = ut + \frac{1}{2} gt {}^{2} \\ = > 40 = 10 \times t + \frac{1}{2} \times 10 \times t {}^{2} \\ = > 40 = 10t + 5t { }^{2} \\ = > 5t {}^{2} + 10t - 40 = 0 \\ = > 5.(t {}^{2} + 2t - 8) = 0 \\ = > 5.(t {}^{2} - 2t + 4t - 8) = 0 \\ = > 5.(t.(t - 2) + 4.(t - 2) = 0 \\ = > 5.(t + 4).(t - 2) = 0 \\ = > (t + 4)(t - 2) = 0 \times 5 \\ = > (t + 4)(t - 2) = 0 \\ On \: further \: solving, \\ \: we \: get \: the \: value \: of \: t : \\ > > t + 4 = 0 \\ t= \: 0 - 4 = - 4 \\ > > t - 2 = 0 \\t = 0 + 2 = 2 \:$

$Maximum \: time \: to \: reach \: the \: \\ ground = > 2 \: seconds \\ (Since \: time \: can't \: be \: taken \: as \: \\ negative \: value ,\: it \: will \: always \: be \\ \: counted \: on \: positive \: value)$