Question

If a ball thrown in the vertical direction attains a maximum height of 16m, at what height will its velocity be half that of its initial velocity?​

Answer 1

the answer is given below:-

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Answer 2

Given,

Height(s)=16m

g=10m/s²

v=0

u=?

Let us consider a figure,

At C,The velocity becomes 1/2 the initial velocity

In CB,

v=1/2u

Now,

2as=v²-u²

$2 \times ( - 10) \times x = ( \frac{1}{2}u ) ^{2} - {u}^{2} \\ = > - 20x = - \frac{3}{2} {u}^{2} \\ = > 40x = 3 {u}^{2} \\ = > x = \frac{3 {u}^{2} }{40} \: .....(i)$

Again,

2as=v²-u²,Taking AC

$2 \times ( - 10) \times (16 - x) = {v}^{2} - {u}^{2} \\ = > - 20(16 - x) = - \frac{1}{4} {u}^{2} \\ = > {u}^{2} = 80(16 - x)$

By substituting value in (i)

$x = \frac{3 \times 80(16 - x)}{40} \\ = > x = 3 \times 2(16 - x) \\ = > x = 3 \times( 32 -2x) \\ = > x = 96 - 6x \\ = > 7x = 96 \\ = > x = 13.71428..... \\ = > x = 13.7(approx)$

The height at which velocity will become half the initial velocity is 13.7m below the ground and 2.3m above the ground