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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 ft 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).

Respuesta :

Edufirst
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

johanrusli

The maximum height of the baseball is about 90 feet

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

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

[tex]s = -2.7t^2 + 30t + 6.5[/tex]

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

[tex]v = \frac{ds}{dt}[/tex]

[tex]v = -2(2.7)t^{2-1} + 30[/tex]

[tex]\boxed {v = -5.4t + 30}[/tex]

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

[tex]v = -5.4t + 30[/tex]

[tex]0 = -5.4t + 30[/tex]

[tex]5.4t = 30[/tex]

[tex]t = 30 \div 5.4[/tex]

[tex]\boxed {t = \frac{50}{9} ~ s}[/tex]

[tex]s = -2.7t^2 + 30t + 6.5[/tex]

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

[tex]s_{max} = \frac{539}{6}[/tex]

[tex]\large {\boxed {s_{max} \approx 90 ~ feet} }[/tex]

Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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

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