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# Dhoni throws his bat vertically upwards with an initial velocity of $26.2\;m/s$ Determine the height to which the bat will rise above its initial height.

$\begin{array}{1 1} 35.0\;m \\ 38\;m \\ 28\;m \\45\;m \end{array}$

The initial velocity $(v_i)$ of the bat is $+26.2\; m/s$.
(The + sign indicates that the initial velocity is an upwards velocity) The $v_f$ value can be inferred to be 0 m/s since the final state is the peak of its trajectory The acceleration (a) of the bat is $-9.8 m/s^2$
$v_f^2=v_i^2+2 *a *d$
'd' stands for the displacement;
the symbol 't' stands for the time;
the symbol 'a' stands for the acceleration of the object;
the symbol $v_i$ stands for the initial velocity value; and the symbol $'v_f'$ stands for the final velocity.
$(0 m/s)^2 =(26.2 m/s)^2 +2 * (-9.8 m/s^2) * d_0 \large\frac{m^2}{s^2} $$=686.44+\large\frac{m^2}{s^2}$$+(-19.6 m/s^2)*d$
$(-19.6 m/s^2) * d= 0 m^2/s^2 -686.44 m^2/s^2$
$(-19.6 m/s^2) * d= -686.44 m^2/s^2$
$d= \large\frac{(-686.44 m^2/s^2)}{(-19.6 m/s^2)}$
$d= 35.0 m$

SINCE the body is projected verticlly upwards,

g = -9.8 m/s2

u=26.2m/sec

v=0m/sec2

by kinematical equations,

v2 - u2 = 2gh

o - [26.2]2 = 2[-9.8][h]

686.44= 19.6h

h= 686.44/19.6

h=35.2

=> h= 35.5 [approximately]