A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.860 m/s at an angle of 33.6° above the table, and it lands on the magazine 0.0790 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.

Answer 1

Y = ver. = 0.860sin33.6 = 0.4759m/s.

d = Vi*t + 0.5gt^2,
d=0.4759*0.0790 + 0.5*(-9.8)(0.0790)^2,
d = 0.037596 - 0.0305809,
d = 7.02*10^-3m = 7.02mm thick.