Question

A hiker stands on the edge of a cliff 490m above the ground throws a stone horizontally with an initial speed of 15m/s. Neglecting air resistance , find the time taken by the stone to reach the ground and the speed with which it hits the ground.​

Answer 1

Answer:

Explanation:

Take height from ground as distance S = 490m.

Initial Velocity u = 15m/s

acceleration a = gravity = 10m/s(approx.)

S = ut + 1/2a$t^{2}$ (where time is "t" and is not found)

490 = 15t + 1/2*10$t^{2}$

980 = 30t + 10$t^{2}$

30t + 10$t^{2}$ - 980 = 0

Using sridharacharyas formula =

t = (-10 (+/-) $\sqrt{100 - (-29400)}$)/2

t = 80.9 or 90.8  (done using approximate values)

final velocity v = u +at = 15 + 908 = 923 m/s