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Two particles are projected in air with speed $v_{0}$ , at angles $\theta _{1}$ and $\theta _{2}$ to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices

a) angle of project: $q_{1} > q_{2}$

b) time of flight: $T_{1} > T_{2}$

c) horizontal range: $R_{1} > R_{2}$

d) total energy: $U_{1} > U_{2}$

Answers (1)

The answer is the option (a) and (b)

Explanation:

According to formula. Max height of a projectile is

$H = \frac{u^{2} \sin^{2} {\theta }}{2g}$

Option a : $H1>H2$

$\sin ^{2}\theta 1>\sin ^{2}\theta 2$

$(\sin \theta 1+\sin \theta 2)(\sin \theta 1-\sin \theta 2)>0$

So, $(\sin \theta 1+\sin \theta 2)>0 \; or(\sin \theta 1-\sin \theta 2)>0$

So, $\theta 1>\theta 2$ and $\theta$ lies etween 0 and 90 degree i.e. acute

Option b :

$\frac{T1}{T2}=\frac{\sin \theta 1}{\sin \theta 2}$

$T1\sin \theta 2=T2\sin \theta 1$

Since $\sin \theta 1>\sin \theta 2$

$T1>T2$

Posted by

infoexpert24

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