Question

Explain projectile motion and obtain expression for 5

(a) Equation of path of a projectile.

(b) Time of maximum height.

(c) Maximum height of a projectile.

(d) Horizontal range of a projectile.​

Answer 1

Answer:

1st =Trajectory of a projectile is a parabolic trajectory. This explanation and the equation are very useful to class 11 students and IIT JEE aspirants. During the derivation we found, if a projectile is thrown with sped 'u' and at an angle θ with the horizontal, y = x tanθ / ( 2 u^2 cosθ^2)

2nd = The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 ⁡ θ i 2 g . The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin ⁡ .

3rd =The maximum height of the projectile is when the projectile reaches zero vertical velocity. From this point the vertical component of the velocity vector will point downwards. The horizontal displacement of the projectile is called the range of the projectile and depends on the initial velocity of the object.

4th =A projectile is an object that we give an initial velocity, and the gravity acts on it. A projectile's horizontal range is the distance along the horizontal plane. Moreover, it would travel before it reaches the same vertical position as it started from.

Explanation:

Hope it helps you.....