The Astronaut on the moon question

Question

An astronaut on the moon throws a baseball upward. The astronaut is 6​ ft, 6 in.​ tall, and the initial velocity of the ball is 40 ft. per sec. The height s of the ball is given by the equation s=-2.7t²+40t+6.5, where t is the number of seconds after the ball was thrown.

a- After how many seconds is the ball 22

ft above the​ moon's surface?

b- how many seconds will it take for the ball to touch the moons surface

Answer 1

s = -2.7t^2 + 40t + 6.5

s(t) = 22 = -2.7t^2 +40t + 6.5

2.7t^2 -40t + 15.5 = 0

use the quadratic formula-2.

t = 40/5.4 + or - (1/5.4)sqr(1600 -4(2.7)(15.5))

t = (40 +/- sqr(1432.6))/5.4 = (40+37.85)/5.4 or (40-37.85)/5.4 = 77.85/5.4 or 2.15/5.4 = 14.42, 0.40

at time .4 seconds or 14.42 seconds, it's 22 feet high

to find time when it hits the moon's surface, set s = 0 and solve the same way

s = 0 = -2.7t^2 +40t + 6.5

2.7t^2 -40t -6.5 = 0

use the quadratic formula

t = 40/5.4 + or - (1/5.4)sqr(1600 +4(2.7)(6.5))

t =[40 + or - sqr(1670.2)]/5.4

= (40 + or - 40.87)/5.4

=80.87/5.4 (ignore the negative solution)

= 14.98 seconds to hit the moon's surface