Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
0 votes

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} $

Can you answer this question?

2 Answers

0 votes
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$
$(-19.6 m/s^2) * d= 0 m^2/s^2 -686.44 m^2/s^2$
$(-19.6 m/s^2) * d= -686.44 m^2/s^2$
$d= \large\frac{(-686.44 m^2/s^2)}{(-19.6 m/s^2)}$
$d= 35.0 m$
answered Aug 20, 2014 by meena.p
0 votes
SINCE the body is projected verticlly upwards,

g = -9.8 m/s2



by kinematical equations,

v2 - u2 = 2gh

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

686.44= 19.6h

h= 686.44/19.6


=> h= 35.5 [approximately]
answered Aug 14, 2016 by donkadasnd

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App