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dexar [7]
3 years ago
14

An astronaut on the moon throws a baseball upward. the astronaut is 6ft, 6 in. tall and the initial velocity of the ball is 30 f

t per second. the height s of the ball in feet is given by the equation 2.7 30 6.5, 2 s ï½ ï­ t ï« t ï« where t is the number of seconds after the ball is thrown. (note: see where the initial height and initial velocity values play a role in the equation).

Physics
2 answers:
Ainat [17]3 years ago
5 0
The equation is garbled and the question is missing.

I found this equation for the same statement:

S = - 2.7t ^2 + 30t + 6.5

And one question is: after how many seconds is the ball 12 feet above the moon's surface?

Given that S is the height of the ball, you just have to replace S with 12 and solve for t.

=> 12 = - 2.7 t^2 + 30t + 6.5

=> 2.7t^2 - 30t - 6.5 + 12 = 0

=> 2.7t^2 - 30t + 5.5 = 0

Now you can use the quadratic equation fo find t:

t = { 30 +/- √ [30^2) - 4(2.7)(5.5)] } / (2*2.7)

=> t = 0.186s and t = 10.925 s

Answer: after 0.186 s the ball is at 12 feet over the surface, and again 10.925 s

katovenus [111]3 years ago
3 0

The maximum height of the baseball is about 90 feet

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

The height s of the ball in feet is given by the equation:

s = -2.7t^2 + 30t + 6.5

To find the velocity function, the above equation will be derived as follows:

v = \frac{ds}{dt}

v = -2(2.7)t^{2-1} + 30

\boxed {v = -5.4t + 30}

At the maximum height, the speed is 0 m/s , then:

v = -5.4t + 30

0 = -5.4t + 30

5.4t = 30

t = 30 \div 5.4

\boxed {t = \frac{50}{9} ~ s}

s = -2.7t^2 + 30t + 6.5

s_{max} = -2.7(\frac{50}{9})^2 + 30(\frac{50}{9}) + 6.5

s_{max} = \frac{539}{6}

\large {\boxed {s_{max} \approx 90 ~ feet} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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Read 2 more answers
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0
3 years ago
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