Question

A ball is projected vertically upwards from ground with velocity 12 m/s in presence of air drag whose
magnitude always remains constant during the motion. If ball takes 1 second to reach the hiohes
point in presence of air friction then calculate the difference of time of flights of the ball in presence
and in absence of air friction. (g = 10 ms?)
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Answer 1

Figure:

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(0,0)(0,-40){2}{\circle{10}}\multiput(-8,0)(16,0){2}{\vector(0,-1){10}}\multiput(0,5)(0,-40){2}{\vector(0,1){10}}\multiput(0,-45)(8,10){2}{\vector(0,-1){10}}\multiput(1,14.5)(0,-70){2}{\sf{v}}\multiput(-7,-11)(8,-16){2}{ \sf{a_d} }\multiput(7,-13)(0,-35){2}{\sf{g}}\end{picture}$

Solution:

Time of flight in absence of air :

$T_{0}$ = $\sf{\dfrac{2u}{g}}$

$\longrightarrow$ $T_{0}$ = $\sf{\dfrac{2 \times 12}{10}}$

$\longrightarrow$ $T_{0}$ = 2.4 s

Time lf flight in presence of air :

$\sf{a_{net}}$ = $\sf{a_{d} + g}$

Now, $\sf{T_{a,u} = \dfrac{u}{g + a} \implies g + a_{d} = 12}$

$\sf{a_{d} = 2 m/s^{2}}$

S = $\dfrac{1}{2}$ a't'²

$\longrightarrow$ 6 = $\dfrac{1}{2}$ × (10 - 2) t'²

$\longrightarrow$ $\dfrac{12}{8}$ = t'²

$\longrightarrow$ t' = $\sqrt{\dfrac{3}{2}}$

$\longrightarrow$ t' = 1.22 s

Difference $\Delta\:t$ = [2.4 - (1 + 1.22)]

$\longrightarrow$ $\sf{\blue{\Delta t\:=0.18s}}$

Hence, difference of time of flights of the ball in presence and absence of air friction is 0.18s

Answer 2

Answer:

Hope this helps!!!!!!!