the above problem can be solved as

horizontal distance to be travelled is $S_y$ 100 - 92= 8m

vertical distance to be travelled $S_x$ = 16m

let $v_x$ be the horizontal velocity =constant here

now vertical sistance equation is $S_y$=ut+$\frac{1}{2}$ g$t^2$

where u is vertical initial velocity

t is time taken to travel 8m i.e., reaching time to land second hill

as vertical initial velocity is 0

so above equation can be written as for time t,

 t=$\sqrt{\frac{2X{S}_y}{g}}$

so t= $\sqrt{\frac{2x8}{9.8}}$ sec.

so t = 1.277 sec.

now taking horizontal distance as $S_x$= $v_x$ X t

16= $v_x X 1.277$

on solving above we get minimum horizontal velocity required to land on second hill as $V_x$= 12.52 m/sec

firstly we calculate the base diagonal

so on applying pythagorus theorem, we get the diagonal of base

D₁=$\sqrt{7^2+4^2}$=$\sqrt{65}$

now joining the point from which mosquito enters and it sits

we will found a diagonal which is required displacement made by him

so Displacement made by mosquito = $\sqrt{{D^2}_1+3^2}$

required displacement=$\sqrt{74}$m

2. For the displacement vector $\stackrel{\rightharpoonup }{d}=$(x₂-x₁)i+(y₂-y₁)j+(z₂-z₁)k

so required vctor id given by $\overrightarrow{d}$=7i+4j+3k

suppose initially the distance between the gaurd of A and driver of B be x

initial speed of trains = 72 km/hr = 200m/s

Distance travelled by train A in 50 sec is S₁= 200 X 50=10000m

Distance travelled by train B in 50 sec is S₂= 200X50+(1/2)1X50²=11250m

for the initial distance x between trains

                 x+400+400=11250-10000

                   x+800=1250

                  x=1250-800

              x=450m

hence the initial distance between the guard of A and driver of B is 450m

Linear motion :-  A type of motion in which particle moves in straight line. It follows all linear kinematic equations like v = u + at, S = ut + 1/2at² , v² = u² + 2aS etc.
Example :- motion of bikes, walking man, running athletic etc are example of linear motion.
Rotational motion :-  A motion of rigid body which takes place in such a way that all particles of its moves in a circle about an axis with a common angular velocity.
Example :- motion of stone, rotation of earth, planets are the example of rotational motion. It follows both linear as well as angular equations. angular equations like ω = ω₀ + αt etc

Speed of the aircraft, v = 720 km/h = 720 × 5 / 18  =  200 m/s, Acceleration due to gravity, g = 10 m/s²
Angle of banking, θ = 15°
For radius r, of the loop, we have the relation:
tan θ = v² / rg
r = v² / g tan θ
= 200² / (10 × tan 15)
= 4000 / 0.26
= 14925.37 m= 14.92 km

Two vectors will be maximum when angle between them is 0 degree

Two vectors will be minimum when angle between them is 180 degree

applied force = opposite force causing the body to be stop  i.e., terminal velocity v=0m/s

s=ut+$\frac{1}{2}$at²

0.8=ux5+$\frac{1}{2}$ax5²

0.8=5u+25a/2

1.6=10u+25a:.........................eq.1

now acceleration a=$\frac{0-u}{5}$

u+5a=0:.....................................eq.2

on solving eq.1 and eq.2 we get

a= -0.064m$s^{-2}$

force=ma

so required force =10x0.064

                            =0.64N

An athlete always runs for some distance before taking a jump so that inertia of motion may help him in his muscular efferts to take a longer jump.

let two forces be 3x and 5x

now for magnitude of resultant force F=$\sqrt{(3x{)}^2+(5x{)}^2+2\left(3x\right)\left(5x\right)\cos 60}$

on solving we get x=5N

therefore required forces are 15N and 25N

distance travelled by the ball between is final height and ground

H+h=0x6+$\frac{1}{2}$9.8x6²          where h is the height of the tower and H is the height between top of the tower and final point in the air

H+h=176.4m...................eq1

now when it started going upward, according tonewtons equation of motion

v²=u² -2gH

0=19²-2x9.8H

H=18.418m

now on putting the value of H in eq.1

we get required height of tower h=157.982m

Gravitational forece is called as conservative force because it acts equally to all objectes irrespective of thier mass and size if placed at equal distance.

Path lenght is the total distance travelled by an object from the starting point regardless of where it travelled .

In physics, there are two definitions for "path length." The first is defined as the total distance an object travels. Unlike displacement, which is the total distance an object travels from a starting point, path length is the total distance travelled, regardless of where it travelled. The second is synonymous with wavelength and is used in calculating constructive and destructive interference of waves.