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2.Motion in Straight Line
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A player throws a ball upwards with an initial speed of $29.4\; m s^{-1}$. To what height does the ball rise and after how long (in second) does the ball return to the player's hands $?$ (Take $g=9.8 \;m s^{-2}$ and neglect air resistance).
A
$22.1\;m\;\;and \;\;6\;s$
B
$32.1\;m\;\;and \;\;5\;s$
C
$44.1\;m\;\;and \;\;6\;s$
D
$44.1\;m\;\;and \;\;8\;s$
Solution
From third equation of motion, height ( $s$ ) can be calculated as:
$v^{2}-u^{2}=2 g s$
$s=\frac{v^{2}-u^{2}}{2 g}$
$=\frac{(0)^{2}-(29.4)^{2}}{2 \times(-9.8)}=44.1 m$
From first equation of motion, time of ascent ( $t$ ) is given as:
$v=u+{a t}$
$t=\frac{v-u}{a}=\frac{-29.4}{-9.8}=3 s$
Time of ascent $=$ Time of descent Hence, the total time taken by the ball to return to the player's hands $=3+3=6$ s.
Standard 11
Physics
