Answer:
42.7m
Explanation:
Given:
h'= 1.8 m
r= 1.5 m
ω = 3 rev/s
ω = 3 ₓ 2 π rad/s =>6 π rad/s
Speed of the ball 'V' when it releases from the rope is given by,
V= ω x r
V= 6 π x 1.5
V=28.3 m/s
As we know that acceleration due to gravity always in downward direction.
[tex]V_f^2=V_o^2+2as[/tex]
The final speed will be zero
[tex]0=28.3^2 - 2(9.8)(h)[/tex]
h = 40.8 m
So the height from the ground,H = h' + h
H= 40.8 + 1.8 =42.7m
Answer:
The maximum height is [tex]H =80 \ m[/tex]
Explanation:
From the question we are told that
The length of the rope is [tex]r = 1.50 \ m[/tex]
The rate at which the ball was whirled is [tex]R = 3 rps[/tex]
The height above the ground when released is [tex]h = 1.8 \ m[/tex]
The initial velocity of the ball is mathematically represented as
[tex]v = \frac{d}{t}[/tex]
Where d is the distance which is the circumference of the circular which can be calculated as
[tex]d = 2 \pi r[/tex]
[tex]d = 2 * 3.143 * 1.5[/tex]
[tex]d =9.426 \ m[/tex]
Now the time is 1 second as stated in the question so
[tex]v =\frac{9.426}{1}[/tex]
[tex]v =9.426 \m/s[/tex]
From the law of energy conservation
[tex]PE + KE = PE_{max}[/tex]
Where PE is the potential energy at the first height stated in the question
Which is mathematically represented as
[tex]PE = mgh[/tex]
KE is the kinetic energy at the first level which is mathematically represented as
[tex]KE = \frac{1}{2} m v^2[/tex]
[tex]PE_{max}[/tex] is the potential energy at the maximum height which is mathematically represented as
[tex]PE_{max} = mgH[/tex]
Where H is the maximum height
So
[tex]mgh + \frac{1}{2} mv^2 = mgH[/tex]
[tex]gh + \frac{1}{2} v^2 = gH[/tex]
[tex]H = \frac{gh + \frac{1}{2} v^2 }{g}[/tex]
[tex]H = \frac{(9.8 ) * 1.8 + \frac{1}{2} (9.426)^2 }{9.8}[/tex]
[tex]H =80 \ m[/tex]
