Respuesta :
Answer:
Step-by-step explanation:
Given
height of building h=116 ft
Two textbooks are thrown from building i.e. one is dropped and another is thrown with a velocity of 48 ft/s
for first textbook let [tex]t_1[/tex] be the time required to reach ground
[tex]h=0+\frac{gt_1^2}{2}[/tex]
[tex]116=\frac{32.2\times t_1^2}{2}[/tex]
[tex]t_1=\sqrt{\frac{233}{32.2}}=2.68 s[/tex]
For second book
[tex]h=ut+\frac{gt_2^2}{2}[/tex]
[tex]116=48\times t_2+\frac{32.2\times t_2^2}{2}[/tex]
[tex]32.2t_2^2+96t_2-233=0[/tex]
[tex]t_2=\frac{-96\pm \sqrt{(96)^2+4\times 32.2\times 233}}{2\times 32.2}[/tex]
[tex]t_2=1.58 s[/tex]
