The velocity is 14 m/s
The parameters given on the question are
mass= 0.060 kg
kinetic energy= 5.9 joules
K.E= 1/2mv²
5.9= 1/2 × 0.060 × v²
5.9= 0.5 × 0.060v²
5.9= 003v²
v²= 5.9/0.03
v²= 196.66
v= √196.66
v= 14 m/s
Hence the velocity of the egg before it strikes the ground is 14 m/s
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answer: in pic
explanation: found it here http://gaaq.weebly.com/uploads/7/8/8/4/78844490/chapter_11_q_as.pdf
Answer:
Maximum height of rocket = 2538.74 m
Explanation:
We have equation of motion s = ut + 0.5 at²
For first 5 seconds
s = 0 x 5 + 0.5 x 40 x 5² = 500 m
Now let us find out time after 5 seconds rocket move upward.
We have the equation of motion v = u + at
After 5 seconds velocity of rocket
v = 0 + 40 x 5 = 200 m/s
After 5 seconds the velocity reduces 9.8m/s per second due to gravity.
Time of flying after 5 seconds

Distance traveled in this 20.38 s
s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m
Maximum height of rocket = 500 +2038.74 = 2538.74 m