Question

Show that if for a given angle of projection the horizontal range is doubled the time of flight become 1.41 times

Answer 1

We have, horizontal range,

R = u² sin(2θ) / g

Here θ is the angle of projection and it is fixed.

So, since θ and g are constants, to double the value of R, the value of u, initial velocity, should be changed. So,

2R = 2 u² sin(2θ) / g

2R = (u√2)² sin(2θ) / g

Let u√2 = u' and 2R = R'. So,

R' = (u')² sin(2θ) / g

So we have had a new initial velocity u√2.

Earlier, the time of flight was,

T = 2 u sinθ / g

Now the time of flight will be,

T' = 2√2 u sinθ / g

T' = √2 · 2 u sinθ / g

I.e., T' = T√2 ≈ 1.41 T

Hence time of flight will be approximately 1.41 times of the original.

Hence the Proof!