A body is dropped from the top of a tower. It acquires a velocity of 2

1.	A body is dropped from the top of a tower. It acquires a velocity of 20 m/s on reaching the ground. Calculate the height of the tower. (Take g = 10 m/s2)
Answer» Given:- Initial velocity(u) = 0m/s Final velocity(v) = 20m/s Gravitational force (g) = 10m/s² To find:- Height of the tower. Solution :- GRAVITY is CAUSE of an object to fall toward the GROUND surface at a faster velocity, the longer the object falls. There is a relation between velocity and gravity, 2AS = v² - u², where ACCELERATION(a) = gravity(g) and s = h(height). ✏ From 3rd equation of motion, $\sf \implies \: 2gh = v {}^{2} - u {}^{2}$ $\sf \implies \: 2 \times 10 \times h = (20) {}^{2} - (0) {}^{2}$ $\sf \implies \: 20h = 400$ $\sf \implies \: h = \dfrac{400}{20}$ $\large \implies \boxed {\boxed {\tt \purple { h = 20 m }}}$ Therefore, height of tower is 20m.