${\mathrm{h}=\mathrm{s}=\frac{1}{2} \mathrm{g} \sin \theta \mathrm{t}^{2}}$
${\text { or } \mathrm{t}=\mathrm{t_s}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}}$
${\text { and } \mathrm{u}_{\mathrm{s}}=\sqrt{2 \mathrm{gh}}=\sqrt{2(\mathrm{g} \sin \theta) \mathrm{s}}}$
while in case of free fall,
$\mathrm{t}_{\mathrm{F}}=\sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}$ and $\mathrm{u}_{\mathrm{F}}=\sqrt{2 \mathrm{gh}}$
$\frac{\mathrm{t}_{\mathrm{F}}}{\mathrm{t}_{\mathrm{s}}}=\sin \theta<1 \mathrm{i.e.}, \mathrm{t}_{\mathrm{F}}<\mathrm{t}_{\mathrm{s}}$
i.e., falling body reaches the ground first.
Also $\quad \frac{\mathrm{u}_{\mathrm{F}}}{\mathrm{u}_{\mathrm{s}}}=1,$ i.e., $\mathrm{u}_{\mathrm{F}}=\mathrm{u}_{\mathrm{s}}$
i.e., both reach the ground with same speed (not velocity as for falling body direction is vertical while for sliding body along the plane downwards).
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$N$ of electrostatic force due to the other two charges.
$_1{H^2}{ + _1}{H^2}{ \to _1}{H^3} + p$
$_1{H^2}{ + _1}{H^3}{ \to _2}H{e^4} + n$
The masses of the nuclei are as follows :
$M({H^2}) = 2.014\;amu;\;M(p) = 1.007\;amu;$
$M(n) = 1.008\;amu;\;M(H{e^4}) = 4.001\;amu$
If the average power radiated by the star is ${10^{16}}W$,the deuteron supply of the star is exhausted in a time of the order of