Explanation
Given,
Initial velocity, $$u=5\,m{{s}^{-1}}$$
Acceleration, $$a=2\,m{{s}^{-2}}$$
Apply kinematic equation of motion
$$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$
$$ s=5\times 10+\dfrac{1}{2}\times 2\times 10^2=150\,m $$
Hence in first 10 second card travel $$150\,m$$
Initial velocity, $$u=15m{{s}^{-1}}$$
Acceleration, $$a=-10\,m{{s}^{-2}}$$
$$ v=u+at $$
$$ v=15-10\times 2=-5\,m{{s}^{-1}} $$
Hence, velocity after $$2\,\sec $$ is $$5\,m{{s}^{-1}}$$ downward.
Object is free fall from height $$h$$
Initial velocity, $$u=0$$
$$ h=\dfrac{1}{2}g{{t}^{2}}\,\,........\,(1) $$
At half time
$$s=u\dfrac{t}{2}+\dfrac{1}{2}g{{\left( \dfrac{t}{2} \right)}^{2}}=\dfrac{1}{4}\left( \dfrac{1}{2}g{{t}^{2}} \right)=\dfrac{h}{4}$$
Height from ground, $$h-\dfrac{h}{4}=\dfrac{3h}{4}$$meter
Hence, at half time height from ground is $$\dfrac{3h}{4}$$m
suppose $$n - balls \to $$
$$final\,velocity\,,\,$$ $$v = 0$$
$$u = 0$$
$$T = {\dfrac{1} {n}}\sec $$
$$0 = u - g{\dfrac{1}{n}}$$
$${\dfrac{g} {n}} = u$$
$${v^2} = {u^2} - 2gh$$
$$0 = {\dfrac{{g^2}} {{n^2}}} - 2gh$$
$${\dfrac{{g^2}} {{n^2}}} = 2gh$$
$$h = {\dfrac{g} {2{n^2}}}$$
$$n = 5$$
$$ = {\dfrac{10} {2 \times 25}}$$
$$ = {\dfrac{1}{5}}m$$
Given that,
Height $$=h$$
Time $$t=\dfrac{3T}{4}$$
Since, the ball is released from rest, its initial velocity is zero. When released from height h it reaches the ground in time T.
If we apply the second equation of motion
$$ s=ut+\dfrac{1}{2}g{{t}^{2}} $$
$$ h=\dfrac{g{{T}^{2}}}{2} $$
$$ {{T}^{2}}=\dfrac{2h}{g} $$
Now, we apply the same equation to calculate the height covered in time 3T/4
$$ s=g\times \dfrac{{{\left( \dfrac{3T}{4} \right)}^{2}}}{2} $$
$$ s=\left( \dfrac{9g}{32} \right)\times {{T}^{2}} $$
$$ s=\left( \dfrac{9g}{32} \right)\times \dfrac{2h}{g} $$
$$ s=\dfrac{9h}{16} $$
Now, the distance from the ground
$$ s=h-\dfrac{9h}{16} $$
$$ s=\dfrac{7h}{16} $$
Hence, the distance from the ground is $$\dfrac{7h}{16}$$
$$\vec{v}=\vec{u}+\vec{a}t$$
$$or\,\,\vec{v}=-\vec{u}-\vec{a}t$$
Direction of initial velocity and direction of acceleration should be same, either negative or positive
Hence, at negative velocity and negative acceleration object will speed up
Time period of free fall from height $$125\,m$$
$$t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times 125}{10}}=5\,\sec $$
$$s=ut+\dfrac{1}{2}a{{t}^{2}}=\dfrac{1}{2}g{{t}^{2}}$$
Ratio of displacement for $${{1}^{st}}$$ sec and last $$1\,\sec $$
$$\dfrac{{{s}_{1}}-{{s}_{0}}}{{{s}_{4}}-{{s}_{5}}}=\dfrac{\dfrac{1}{2}g\left( {{1}^{2}}+0 \right)}{\dfrac{1}{2}g\left( {{5}^{2}}+{{4}^{2}} \right)}=\dfrac{1}{9}$$
Hence , ratio is $$1:9$$
Solution 2
Distance travelled by bus in $$10s \Rightarrow \dfrac {1}{2} \times 1 \times 10^2 = 50 m$$D $$=$$ Distance boy has to cover $$= 10 + 50 = 60 m$$Speed of boy $$=\dfrac {D}{T} = \dfrac {60}{10} = 6\ m/s$$
Velocity attained just before contact:
$$v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 10} = \sqrt {196} = 14\ m/s$$$$u_1 = v$$$$v_1 = u_1 + at$$$$0 = 14 + a (0.029)$$$$a = -(\dfrac {14}{0.029})$$
$$s = ut + \dfrac {1}{2} at^2 = 14 (0.029) + \dfrac {1}{2} (\dfrac {-14}{0.029})(0.029)^2$$$$s = 0.029 (14 + \dfrac {1}{2} (-14))$$
$$ = 0.029 \times 7$$ $$ = 0.203\ m$$ $$= 0.2\ m$$
$$s = ut + \dfrac {1}{2} gt^2$$
$$3 = u (0.5) + \dfrac {1}{2} \times 10 (0.5)^2$$$$3 = (0.5)u + 5(0.25)$$
$$\dfrac {1.75}{0.5} = u$$
$$\Rightarrow u = 3.5\ m/s$$
$$u^2 = u_1^2 + 2gs$$
$$\dfrac {(3.5)^2}{2g} = s$$
$$\Rightarrow s = \dfrac {3.5 \times 3.5}{2g}$$$$\Rightarrow \ s = 0.6125\ m$$
$$3 = u (0.5) + \dfrac {1}{2} \times 10 (0.5)^2$$
$$3 = (0.5) u + 5 (0.25)$$
$$\Rightarrow u = 3.5\ m/s$$$$v = u + gt$$$$v = (3.5) + 10 (0.5) = 8.5\ m/s$$
Please disable the adBlock and continue. Thank you.
