Question

a projectile has its maximum range given by rmax find time to reach maximum hight is √rmax/2g.​

Answer 1

Answer:

A cannon has its maximum range given by Rmax. How can you prove that the height reached in such case is 1/4 Rmax and the time of flight is square root of Rmax/2G?

Can kids learn science at home?

What you are citing in this case is only true when the angle of projection theta is 45 degrees. For other angles of projection it is not true. My proof is found in the next paragraphs.

formula for horizontal range, Rmax = Vi^2 * sin (2 * θ )/g

Formula for maximum height, H = (Vi * sin θ)^2/2g

Equation: Vi^2 * sin (2 * θ )/g = (4 * (Vi * sin θ )^2)/2g

Expanding 4 * (Vi * sin θ )^2)/2g = (4 * Vi^2 * sin^2 θ )/2g

Equating Vi^2 * sin (2 * θ )/g = (4 * Vi^2 * sin^2 θ )/2g

Canceling Vi^2 and g found in both sides

sin (2 * θ ) = 2 * sin^2 θ

sin (2 * θ ) = 2 * sin θ * cos θ

2 * sin^2 θ = 2 * sin θ * cos θ

dividing both sides by 2 and sin θ

sin θ = cos θ and this happens only when the angle is 45 degrees

θ = 45 degrees

Regarding the time of flight t

Let vy = the initial vertical velocity

t = t_rise + t_fall and t_fall = t_rise so t = 2 * t_rise and t_rise = vy/g therefore t = 2vy/g

vy = vi * sin45 so t = 2 * vi * sin45 /g but Rmax = vi^2 * sin (2 * 45 )/g therefore by t = (2vi * √2÷2)/g = (√2 * vi) / g and when compared to √(Rmax)/2g)

(√2 * vi) / g compared to √(Rmax)/2g)

(√2 * vi) / g compared to √(vi^2 * sin (2 * 45 )/g)/2g)

(√2 * vi) / g compared to √(vi^2 * sin (2 * 45 )/0.5)

(√2 * vi) / g compared to √(vi^2 * sin (90)/0.5)

(√2 * vi) / g compared to √(vi^2 * 1/0.5)

(√2 * vi) / g compared to √(vi^2 * 2)

(√2 * vi) / g compared to √2 * vi

The 2 quantities are not equal because the time of flight at 45 degrees is equal to (√2 * vi) / g while the square root of (Rmax)/2g) is equal only to √2 * vi without the divisor g. The earlier value of 2g was canceled.