Explanation:
A diver running 2.6 m/s dives out horizontally from the edge of a vertical cliff and 2.6 s later reaches the water below.
(a) It means in horizontal direction,
Initial velocity is 0 and acceleration is g. Let h is the height of the cliff. Using equation of motion as :
[tex]h=ut+\dfrac{1}{2}gt^2\\\\h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 2.6^2}{2}\\\\h=33.12\ m[/tex]
(b) In horizontal direction,
Initial velocity of the driver is 2.6 m/s and acceleration is equal to 0. Let x is the distance from its base did the diver hit the water.
[tex]d=ut+\dfrac{1}{2}gt^2\\\\d=ut\\\\d=2.6\times 2.6\\\\d=6.76\ m[/tex]
