Question:
Figure (8-E4) shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point $A$, how far away from the track will the particle hit the ground?
Solution:
By conservation of energy of point $A$ and $B$
$\mathrm{mgH}=\frac{\frac{1}{2}}{2} \mathrm{~m} \mathrm{} \mathrm{v}^{2}+\mathrm{mgh}$
$g \times 1=\frac{1}{2} v^{2}+g \times 0.5$
$v=\sqrt{9.8}=0.3 \mathrm{~m} / \mathrm{s}$
After B, body obeys projectile motion, so
$S=u t+\frac{1}{2} a t^{2}$
$-0.5=u \sin \theta t+\frac{1}{2}(-9.8) t^{2}$
$t=0.31 \mathrm{~s}$
and $X=4 \cos \theta \times t$
$X=1$