Question

For any specific velocity of projection,the maximum range s equal to four times of the corresponding height.Discuss...also prove it mathematically​

Answer 1

Explanation:

Motion in A Plane. Prove that the maximum horizontal range is 4 times the maximum height attained by a projectile. The horizontal range is maximum when the angle of projection is 45°. That is, range is 4 times the maximum height attained by a projectile.

Answer 2

Answer:

The range formula is Range = v^2 * sin 2A / g while the maximum height formula is ymax = v^2 * (sin A)^2 / 2g where A = the angle of projection, v is the initial velocity, and g is 9.8 m/s^2.

The condition as stated in the problem is “the maximum range equal to four times of its height”. This happens only when angle A is equal to 45 degrees.

Solving for the range when angle A = 45 degrees

Range = v^2 * sin 2A / g

Range = (v^2 / g) * (sin 2 * 45)

Range = (v^2 / g) * sin 90

Range = (v^2 / g) * 1

Range = (v^2 / g)

Solving for the maximum height when angle A = 45 degrees

ymax = v^2 * (sin A)^2 / 2g

ymax = v^2 * (sin 45)^2) / 2g

ymax = v^2 * [(√2÷2)^2 / 2g

ymax = v^2 * (0.5)/ 2g

ymax = v^2 * 1/2 / 2g

ymax = v^2 / g * (1/4)

Solving for the ratio of range to the maximum height

ratio = range : ymax

ratio = (v^2 / g) : v^2 / g * (1/4)

ratio = 1 : 1/4

At 45 degrees angle of projection the range is four times the maximum height.

Explanation:

1:1/4