A wrecking ball is suspended from a 7.0-m-long cable that makes a 25 ? angle with the vertical. the ball is released and swings down. part a what is the ball's speed at the lowest point? what is the ball's speed at the lowest point? 7.2 m/s 5.1 m/s 3.6 m/s 2.5 m/s

The height of the ball when lifted is given by 7sin(25)=2.96
the gravitational energy is mgh, the kinetic is (1/2)mv². We can set these equal since the pendulum doesn't lose much energy
mgh = (1/2)mv²
we can divide by m (since we don't have it anyways)
gh = v²/2
v=√(gh/2) = √(9.81*2.96/2)=3.8m/s.
Not exactly one of your choices, but the right one none the less

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snehashish65 snehashish65 · Answer 2 · 2021-11-11T12:57:51+01:00

The speed of ball at the lowest point is 7.20 m/s. Hence, option (a) is correct.

Given data:

The length of cable is, L = 7.0 m.

The angle of inclination with vertical is, [tex]\theta = 25^{\circ}[/tex].

The given problem can be resolved by using the conservation of energy. Since, the ball is suspended from a vertically inclined cable. So, the height raised by the ball is,

[tex]h = Lsin \theta\\h = 7.0 \times sin 25^{\circ}\\h=2.96 \;\rm m[/tex]

Now, at lowest point the kinetic energy of ball will be maximum (say KE) while at highest point, the potential energy of ball will be maximum (say PE). Then as per the conservation of energy,

Kinetic energy at bottom = potential energy at highest point

[tex]KE = PE \\\\\dfrac{1}{2}mv^{2}=mgh[/tex]

Here, m is the mass of ball, g is the gravitational acceleration and v is the speed of ball at lowest point.

Solving as,

[tex]\dfrac{1}{2}mv^{2}=mgh\\\\\dfrac{1}{2}v^{2}=gh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2 \times 9.8 \times 2.96}\\\\v \approx 7.20 \;\rm m/s[/tex]

Thus, the speed of ball at the lowest point is 7.20 m/s.

Learn more about the conservation of energy here:

https://brainly.com/question/15707891