Answer:
just before landing the ground
Explanation:
Let the velocity of projection is u and the angle of projection is 30°.
Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.
initial horizontal component of velocity, ux = u Cos 30
initial vertical component of velocity, uy = u Sin 30
Time of flight is given by
![T = \frac{2u Sin\theta }{g}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2u%20Sin%5Ctheta%20%7D%7Bg%7D)
Final horizontal component of velocity, vx = ux = u Cos 30
Let vy is teh final vertical component of velocity.
Use first equation of motion
vy = uy - gT
![v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}](https://tex.z-dn.net/?f=v_%7By%7D%3Du_%7By%7D-%20g%20%5Ctimes%20%5Cfrac%7B2u%20Sin%5Ctheta%20%7D%7Bg%7D)
![v_{y}=u Sin 30 - 2u Sin 30](https://tex.z-dn.net/?f=v_%7By%7D%3Du%20Sin%2030%20-%202u%20Sin%2030)
vy = - u Sin 30
The magnitude of final velocity is given by
![v = \sqrt{v_{x}^{2}+v_{y}^{2}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bv_%7Bx%7D%5E%7B2%7D%2Bv_%7By%7D%5E%7B2%7D%7D)
![v = \sqrt{\left (uCos 30 \right )^{2}+\left (uSin 30 \right )^{2}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cleft%20%28uCos%2030%20%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%28uSin%2030%20%20%5Cright%20%29%5E%7B2%7D%7D)
v = u
Thus, the velocity is same as it just reaches the ground.
Velocity also tells you the direction of motion.
Answer:
Inertia; Newton's First Law
Explanation:
When you hit you're breaks, the force you feel that pulls you forward is due to inertia.
Newton's first law says that "every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force." The question is a perfect example of this.