Explanation
Given that,
Height $$h = 39.2\ m$$
Initial velocity $$u = 0$$
Now, from equation of motion
After crosses half distance, the acceleration due to gravity ceases to act
So,
$$ h=\dfrac{39.2}{2} $$
$$ h=19.6\,m $$
$$ {{v}^{2}}-{{u}^{2}}=2gh $$
$$ {{v}^{2}}=2\times 9.8\times 19.6 $$
$$ v=19.6\,m/s $$
Hence, the velocity is $$19.6\ m/s$$
Given,
Vertical velocity is zero
Horizontal velocity is $$v=15\,m/s$$
Height, $$h=25\,m$$
Time period $$t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times 25}{9.81}}=2.25\ \sec $$
Range, $$R=vt=15\times 2.25=33.75\,m$$
Time $$t=2.5\min $$
Distance $$d=2km$$
Now, firstly for half distance
$$ v=\dfrac{d}{t} $$
$$ t=\dfrac{d}{v} $$
$$ t=\dfrac{60\times 60}{40} $$
$$ t=90\,s $$
Now, total time is
$$ t=150-90 $$
$$ t=60\,s $$
Now, the speed is
$$ v=\dfrac{1\times 60}{1} $$
$$ v=60\,km/h $$
Hence, the speed is $$60 km/h$$
Let total distance covered in the entire journey is $$2x$$.
Let $$t_1$$ is the time to cover half of the distance with a speed of $$6 \ m/s$$.
So,$${{t}_{1}}=\dfrac{x}{6}............(1)$$
Let, for the second part total time is $$t_2$$.
Let the distance covered in first half time with a speed of $$2 \ m/s$$ is $$x_1$$. So, $${{x}_{1}}=\dfrac{{{t}_{2}}}{2}\times 2={{t}_{2}}............(2)$$
For the second half the distance covered with a speed of $$4 \ m/s$$is $$x_2$$. So, $${{x}_{2}}=\dfrac{{{t}_{2}}}{2}\times 4=2{{t}_{2}}..........(3)$$
Now,
$$ {{x}_{1}}+{{x}_{2}}=x $$
$$ {{t}_{2}}+2{{t}_{2}}=x $$
$$ {{t}_{2}}=\dfrac{x}{3}............(3) $$
Hence, $$total \ time = t_1+t_2 = \dfrac{x}{6} + \dfrac{x}{3}$$
Now , $$ average\,\,speed=\dfrac{total\,\,distance}{total\,\,time} $$
$$ s=\dfrac{2x}{\dfrac{x}{6}+\dfrac{x}{3}}=\dfrac{2x}{3x}\times 6 $$
$$ s=4\,m/s $$
Given:
Acceleration $$a=2\,m/{{s}^{2}}$$
Time $$t=5\,s$$
Initial velocity $$u=0$$
We know from the equation of motion,
$$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$
$$\Rightarrow s=0+\dfrac{1}{2}\times 2\times 5\times 5 $$
$$ s=25\,m $$
Hence, the distance moved by the particle is $$25\ m$$
Acceleration of lift $$a=2\,m/{{s}^{2}}$$
We know that,
Acceleration of ball relative to lift = acceleration of ball relative to ground – acceleration of lift with respect to ground
$$ a'=-10\hat{j}-\left( -2\hat{j} \right) $$
$$ a'=8\hat{j} $$
$$ s=ut-\dfrac{1}{2}a{{t}^{2}} $$
$$ s=0+\dfrac{1}{2}\times 8\times {{t}^{2}} $$
$$ {{t}^{2}}=\dfrac{1}{2} $$
$$ t=\dfrac{1}{\sqrt{2}}\,s $$
Hence, the time take after which ball will strike the floor is $$\dfrac{1}{\sqrt{2}}\,s$$
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