Question
A body is thrown horizontally with a velocity $$\sqrt {2gh} $$ from the top of a tower of height $$h.$$ It strikes the level ground through the foot of the tower at a distance $$x$$ from the tower. The value of $$x$$ is
A.
$$h$$
B.
$$\frac{h}{2}$$
C.
$$2h$$
D.
$$\frac{{2h}}{3}$$
Answer :
$$2h$$
Solution :
Using equation to trajectory
$$\eqalign{ & - h = x\tan \left( {{0^ \circ }} \right) - \frac{{g{x^2}}}{{2\left( {2gh} \right)\left( {{{\cos }^2}{0^ \circ }} \right)}} \cr & \Rightarrow x = 2h \cr} $$
Using equation to trajectory
$$\eqalign{ & - h = x\tan \left( {{0^ \circ }} \right) - \frac{{g{x^2}}}{{2\left( {2gh} \right)\left( {{{\cos }^2}{0^ \circ }} \right)}} \cr & \Rightarrow x = 2h \cr} $$
