Answer to Question #242498 in Electric Circuits for lovely

Question

  A stone is released from rest at a height of 122.5 m. a) What is its velocity after 2 seconds? b) How many seconds will it take for the stone to reach the ground, and c) With what velocity will it hit the ground?

 

Answer 1

$h=122.5 \;m \\ g = 9.8 \;m/s^2$

a) t= 2 sec

Distance traveled in 2 sec

$S = \frac{1}{2}gt^2 \\ S = \frac{1}{2} \times 9.8 \times 2^2 = 19.6 \;m \\ v^2 -u^2 = 2gS \\ v^2 -0^2 = 2 \times 9.8 \times 19.6 \\ v = \sqrt{384.16} = 19.6 \;m/s$

b) Using the equation of motion

$h=ut+ \frac{1}{2}gt^2 \\ 122.5 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2 \\ t^2 = \frac{2 \times 122.5}{9.8} \\ t = \sqrt{25} = 5 \;sec$

c)

$v^2=u^2 +2gh \\ v^2 = 0^2 +2gh \\ v= \sqrt{2gh} \\ v= \sqrt{2 \times 9.8 \times 122.5} \\ v = 49 \;m/s$