Question

Would any one derive and explain the formula...
Range of a projectle=2 height of a projectile.
Any help would be greatly appreciated...

Answer 1

@irishboy123

Answer 2

@arindameducationusc

Answer 3

@michele_laino

Answer 4

can you scan or link the question pls :p

Answer 5

range R for generalised velocity v at angle \(\theta \) is given by: \(R = \dfrac{v^2 \sin 2 \theta} {g} \)

the follows from equations of motion

from \(v = u + at\), we have half flight time \(t_{1/2} \) is \(t_{1/2} = \dfrac{0 - v \sin \theta}{-g} = \dfrac{ v \sin \theta}{g}\)

thus horizontal range R is \(R = 2 \times t_{1/2} \times v \cos \theta = 2 \dfrac{ v \sin \theta}{g} v \cos \theta = \dfrac{v^2 \sin 2 \theta} {g}\)

the max height \(H_{max}\) for that generalised trajectory comes from \(v^2 = u^2 + 2ax\) and gives us
\[H_{max} = \dfrac{v^2 \sin^2 \theta}{2g}\]

if you are looking for a trajectory where \(R = 2 H_{max}\) then

\(\dfrac{v^2 \sin 2 \theta} {g} = 2 \dfrac{v^2 \sin^2 \theta}{2g}\)
\(\implies \sin 2 \theta = \sin^2 \theta \)
\(\implies \tan \theta = 2 \)

Answer 6

I agree with @IrishBoy123

The question must be like this

During a projectile motion if the max height equals horizontal rangle, then the angle of projection with the horizontal is?


And @IrishBoy123 has given the answer