First slide

Projection Under uniform Acceleration

Question

A cannon ball has a range R on a horizontal plane. If h and h' are the greatest heights in the two paths for which this is possible, then

Moderate

A

$R = 4 \sqrt{({hh}^{'})}$
B

$R = \frac{4 h}{h^{'}}$
C

$R = 4 {hh}^{'}$
D

$R = \sqrt{{hh}^{'}}$

Solution

The range is same for angles $α$ and $(\frac{π}{2} - α)$ .
The range R for $α$ is given by
$R = \frac{2 u^{2} \sin αcos α}{g}$ ….(1)
Maximum height, $h = \frac{u^{2} \sin^{2} α}{2 g}$ … (2)
The range and height corresponding to angle $(\frac{π}{2} - α)$ are
$R = \frac{2 u^{2} \sin (\frac{π}{2} - α) \cos (\frac{π}{2} - α)}{g}$
and $h^{'} = \frac{u^{2} \sin^{2} (\frac{π}{2} - α)}{2 g} = \frac{u^{2} \cos^{2} α}{2 g}$
Now $h \times h^{'} = \frac{u^{4} \sin^{2} α \times \cos^{2} α}{4 g^{2}}$
$\begin{array}{l} \sqrt{{hh}^{'}} = \frac{u^{2} \sin αcos α}{2 g} = \frac{R}{4} \\ ∴ R = 4 \sqrt{({hh}^{'})} \end{array}$

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