Question

Your friend is standing on a 45-foot balcony. He asks you to throw a baseball up to him. You throw a baseball with an initial upward velocity of 50 feet per second. Assuming that you released the baseball 6 feet above the ground, did it reach your friend? Explain.

USE THE NUMBERS THAT ARE IN THE WORD PROBLEM PLEASE!

Answer 1

Step-by-step explanation:

from grnd to top= 45-6 = 39 ft.

thus, $s = 39 ft.$

$u = 50 ft. / sec$

$g= 10 m/s^2$≈$33 ft. / sec$

By using 3rd eq. of motion,

$v^2-u^2 = 2as$        $v= \sqrt{2gs+u^2}$$=\sqrt{2*33*39+50^2} =\sqrt{2574 + 2500} =\sqrt{5074}$

$v=u+at => t=\frac{v-u}{a} = \frac{71-50}{33} =\frac{21}{33} =\frac{7}{11}$$=0.63$ sec