Question

derive the formula for the range and maximum height attained by a projectile thrown from the origin with initial velocity vector u at angle theta to the horizontal​

Answer 1

Range of projectile:

Maximum distance which projectile covers in horizontal direction is called as range of projectile.

To calculate the range multiply the horizontal component of velocity with the total time.

R = Vx × t

R = Vicosθ ×2 Visinθ/g

R = Vi^2sin2θ/g

The range will be maximum if θ = 45 degrees.

Height of projectile:

To determine the height of projectile third equation of motion is used

2as = Vf^2  - Vi^2  

Final velocity will be zero and intial velocity will be Vi sin θ so,

2(-g)h = -Vi^2 sin^2 θ

h = Vi^2 sin^2 θ/2g

if θ = 90 degrees then  

hmax = Vi^2/2g