Question

derive the formula for time take to achieve maximum, total time of flight, maximum height and range of projective​

Answer 1

Answer:

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 ⁡ .

To find the time of flight, determine the time the projectile takes to reach maximum height. The time of flight is just double the maximum-height time. At maximum height, vy = 0. The time of flight is also determined solely by the initial velocity in the y direction and the acceleration due to gravity.

R = Range of projectile t = time required to attain max. height, T = time of flight ... Substituting values of equation (1) in (2) we get. Projectile ... The time taken by the body to reach the maximum height is called time of ascent.

Horizontal distance traveled can be expressed as x = Vx * t , where t is the time.

Vertical distance from the ground is described by the formula y = h + Vy * t – g * t² / 2 , where g is the gravity acceleration.

Answer 2

Answer:

• at max. height

vertical velocity will become 0

• horizontal velocity will remain constant because their is no acceleration in horizontal direction
• time of flight will be twice the time taken to reach maximum height from point of projection.