Considering projectile motion, we cannot say that the motion of an object remains rectilinear under constant magnitude of acceleration and its direction.

In a projectile motion, the acceleration is constant and equal to 'g' which is always directed towards the centre of the Earth. However, when it comes to the path followed by the projectile, it is parabolic in nature.

Time of flight(T)=(2H/g)

Range(R)=u*T

R=430*1000/3600*(2*1200/9.8)

R=1869.21m

Now,

tan(theta)=H/R=0.64

Theta=32.62

Since the horizontal component of the velocity u of such a projectile is constant, there is no acceleration in this direction. So the acceleration of the projectile is taking place in the vertical direction. The acceleration is parallel or anti-parallel to the vertical component of u and hence there is no point at which the acceleration is parallel to u. At the maximum height, the vertical component of u is zero but u is horizontal. So the acceleration is perpendicular to u at the maximum...

The acceleration of an object refers to the change of velocity at a certain rate including both slowing down and speeding up.
(i) So it is possible for the object to be slowing down while its acceleration is increasing in magnitude. In such a case, the object suffers an extreme deceleration and comes to rest rapidly.
(ii) It is also possible for the object to be speeding up while its acceleration is decreasing. In such a case, the acceleration goes on decreasing to become zero and finally the...

Solution

Velocity of car A = slope of x-t of A = 14-6/1.6 = 5km/hs

Velocity of car B = slope of x-t of B =14-0/1.4 = 10km/hs

Velocity of car C = slope of x-t of c = 14-2/1 =12km/hs

A) Car C has the highest speed and car A has the lowest.

B) There is no point where all three line intersect. so, all three car never meet at a same position.

C) When car A passes car C, car B is at 6km.

D) Time interval= 1.2-0.6 = 0.6hr

E) Relative velocity of Car C with respect to car A(v_CA)=V_C-V_A=12-5 =7km/hr

F) R...

Initial velocity(u)= 0m/s

Final velocity(v)= 2.2m/s

Displacement(d)= 24cm=0.24m

Now, v²=u²+2as

(2.2)² = 0²+2a0.24

a = 4.840.48

a = 10.1Ansm/s²_Ans

1. Given: u = 20,
   s = -5 t = ?,
   a = -10
   We know,
   s = ut + 1/2 at²
   -5= 20t - 1/2 x 10 x t²
   -5 = 20t - 5t²
   5t² - 20t - 5 = 0
   t = 4.2s or t = -0.2s
   Time of flight = 4.2s
2. Given: u = 20,
   a = -10,
   v = 0,
   s = ?
   We have,
   v² = u² + 2as
   0 = 400 - 20s
   s = 20m
   Maximum height = 5 + 20 = 25m
3. Given: u = 5,
   a = 0,
   t = 4.2
   s = 5 x 4.2sec
   s = 21m

1. Newton's First Law of Motion (Law of Inertia):

- "An object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an external force."

- In simpler terms, this means that an object will maintain its state of motion (whether at rest or moving) unless an external force acts upon it. If no force is applied, the object will continue its motion or remain at rest.

2. Newton's Second Law of Motion (Law of...