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pshichka [43]
3 years ago
11

Provides some background for this problem. a ball is thrown vertically upward, which is the positive direction. a little later i

t returns to its point of release. the ball is in the air for a total time of 7.70 s. what is its initial velocity? neglect air resistance.
Physics
1 answer:
julsineya [31]3 years ago
6 0
For an object thrown vertically upward, its time of flight is equal to the equation below:

t = 2v/g
where
t is the time of flight
v is the initial velocity
g is the acceleration due to gravity equal to 9.81 m/s²

Substituting the values,

7.07 = 2v/9.81
v = 34.68 m/s
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A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constan
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Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, \theta=12\ rad

Final angular speed, \omega_f=17\ rad/s

The acceleration of the CD,\alpha =3\ rad/s^2

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta

Put all the values,

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3 0
3 years ago
A basketball player throws the ball at a 47 angle above the horizontal to a hoop which is located a horizontal distance L = 5.0
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Answer:

v_0 =1.71

Explanation:

the parabolic movment is described by the following equation:

y = tan(a)x-\frac{1}{2v_0^2(cos(a))^2}gx^2

where y is the height of the ball, a is the angle of launch, v_0 the initial velocity, g the gravity and x is the horizontal distance of the ball.

So, if we want that the ball reach the hood, we will replace values on the equation as:

0.8 = tan(47)(5)-\frac{1}{2v_0^2(cos(47))^2}(9.8)(5)^2

Finally, solving for v_0, we get:

v_0=\sqrt{\frac{-9.8(5)^2}{(0.8-tan(47)(5))2cos^2(47)}}

v_0 =1.71

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