Question #64bde

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Answer 1 · 2017-08-22T17:20:41

Option $(2)$ $(2)$ is correct.

Range $R$ $R$ of a projectile is given by

$R = \frac{v_{0}^{2} sin 2 θ}{g}$ $R=(v_0^2sin2theta)/g$
For above, for a given angle and constant value of $g$ $g$ we conclude that:

$\Rightarrow R \propto v_{0}^{2}$ $=>Rpropv_0^2$

Now, if velocity $v^{2}$ $v^2$ is double of $v_{0}$ $v_0$ , then range $R'$ becomes $4$ times of range $R$ , i.e;

$R'=4R$

Four times means $400%$ ,

$=>R'=400%R$

Now, percentage change in range can be calculated as

Change $=R'-R=400%R-100%R=300%R$

$implies$ percentage change is $300%$ .

Hecne, option $(2)$ is correct.