A rock is thrown downward from an unknown height above the ground with an initial speed of 10m/s. It strikes the ground 3s later
1 answer:
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Answer:
Range of arrow = 225.09 meter
Final horizontal velocity = 34.47 m/s
Explanation:
We have equation of motion , where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.
Considering the vertical motion of arrow ( up direction as positive)
We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = .
t = 6.53 seconds or t = -0.63 seconds
So time = 6.53 seconds.
Considering the horizontal motion of arrow
u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a =
So range of arrow = 225.09 meter
Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The velocity is
Explanation:
The kinetic energy of the 9 kg can be determined by these expression
Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction
Now to obtain the initial energy stored
Let U denote the initial energy stored and
Where x is the length the spring is displaced
k is the force constant of the string
Now referring to the formula above
i.e Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction
and we are told that coefficient of friction = 0.4 and the mass is 9 kg ,the acceleration due to gravity this displacement length of spring = 0.6
Therefore
