Question #242498

  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?

 



1
Expert's answer
2021-09-26T18:47:43-0400

h=122.5  mg=9.8  m/s2h=122.5 \;m \\ g = 9.8 \;m/s^2

a) t= 2 sec

Distance traveled in 2 sec

S=12gt2S=12×9.8×22=19.6  mv2u2=2gSv202=2×9.8×19.6v=384.16=19.6  m/sS = \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+12gt2122.5=0×t+12×9.8×t2t2=2×122.59.8t=25=5  sech=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)

v2=u2+2ghv2=02+2ghv=2ghv=2×9.8×122.5v=49  m/sv^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


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