Question

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Answer 1

hope it helps you......

Answer 2

$\huge\bigstar\mathfrak\black{\underline{\underline{SOLUTION:}}}$

The stone is thrown upward from height,h= 85m from the ground.

If reaches the ground in 5sec.

Let the initial velocity be 'u'.

The displacement is '-85m' & acceleration is '-10m/s^2 as they are downward so the negative sign comes.

Using,

$s = ut + \frac{1}{2} {at}^{2} \\ = > - 85 = 5u - \frac{1}{2} \times 10 \times {5}^{2} \\ = > u = 8m/s$

At the greatest height the ground, the stone comes to momentary rest. That means it's velocity becomes zero.

Using,

${v}^{2} = {u}^{2} + 2as \\ = > 0 = {8}^{2} - 2 \times 10 \times s \\ = > s = 3.2m$

Thus, the maximum height from the ground is 3.2 + 85= 88.2m

Velocity with which it reaches the ground is,

v= u+ at

=) v= 8- 10×5

=) v= 8-50

=) v= -42m/s [downward]

hope it helps ☺️