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).

Question

Rohan Glenn

Physics

16 August, 10:17

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).

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Chessie · Answer 1

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