Answer:
its terminal velocity is 19.70 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Explanation:
Firstly,
given that
m = 580g = 0.58kg
Area A = 0.11 * 0.22 = 0.0242m
g = 9.8
idensity constant p = 1.21 kg/m^3
the terminal velocity of the sphere Vt is ;
Vt = √ ( 2mg / pCA)
we substitute
Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)
Vt = √ (11.368 / 0.029282)
Vt = √ ( 388.22)
Vt = 19.70 m/s
its terminal velocity is 19.70 m/s
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
The Velocity of the person is;
V2 = √ 2ax
V2 = √ ( 2 * 9.8 * 4 )
V2 = √ (78.4)
V2 = 8.85 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
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Hope this helped </span>
Answer:
<em>The velocity of the ball as it hit the ground = 19.799 m/s</em>
Explanation:
Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.
v² = u² + 2gs.......................... Equation 1
Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball
<em>Given: s = 20 m, u = 0 m/s</em>
<em>Constant: g = 9.8 m/s²</em>
<em>Substituting these values into equation 1,</em>
<em>v² = 0 + 2×9.8×20</em>
<em>v² = 392</em>
<em>v = √392</em>
<em>v = 19.799 m/s.</em>
<em>Therefore the velocity of the ball as it hit the ground = 19.799 m/s</em>
Since initial velocity is zero hence , u = 0
=> d = 1/2 * a * t2
on solving we get
d = 86.436 metres
Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2
Answer:
35m/s
Explanation:
150 m/s divided by 4 is 35