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Mademuasel [1]
3 years ago
11

A projectile is launched at an angle above the

Physics
1 answer:
erica [24]3 years ago
7 0
If we neglect friction/air resistance, then the horizontal component
doesn't change, and the vertical component becomes (9.8 m/s downward)
greater each second thanks to gravity.

So, after 2 seconds, the horizontal component is still 40 m/s, and the
vertical component is (30 - 2·9.8) = 10.4 m/s upward.

Choice #1 says this.
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A bowling ball with a circumference of 27 in. weighs 14.8 lb and has a radius of gyration of 3.43 in. If the ball is released wi
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Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, \mu = 0.19

Now,

We know that the circumference of the circle is given by:

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Also, in case of rolling, we know that:

Angular velocity, \omega = \frac{v}{R}

v = \omega R

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

m\omega' = m\omega + I\omega

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v'R = vR + k^{2}\times \frac{v}{R}

276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}

1184.04 - 756.903 = 4.29v

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Now,

Friction force, f = \mu mg

Also, acceleration of the ball can be computed as:

\mu mg = - ma

a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}

Now, the distance, d covered by the ball before rolling without slipping:

v^{2} = v'^{2} + 2ad

99.56^{2} = 276^{2} + 2\times (- 73.357)d

99.56^{2} - 276^{2} = 2\times (- 73.357)d

d = 451.65 in = 37.53 ft

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