A) 6.25m
B) 2.5m
C) 3.75m
D) 5m
E) 1.25m
Correct Answer: B
Solution :
Maximum height of projectile \[h=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\therefore \] \[h=\frac{{{(10)}^{2}}\times {{\sin }^{2}}(30{}^\circ )}{2\times 10}\] \[=\frac{5}{4}=1.25\,m\] Time to reach maximum height \[t=\frac{u\sin \theta }{g}\] \[\therefore \] \[t=\frac{10\times \sin 30{}^\circ }{10}=0.5\,s\] So, distance of vertical fall in 0.5 s \[s=\frac{1}{2}g{{t}^{2}}\] or \[s=\frac{1}{2}\times 10\times {{(0.5)}^{2}}=1.25\,m\] \[\therefore \] Height of second ball \[=1.25+1.25=2.5m\]You need to login to perform this action.
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