A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.890 m/s at an angle of 33.6° above the table, and it lands on the magazine 0.0750 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.

Answer 1

Vo = 0.890 m/s @ 33.6 Deg.
Xo = Hor. = 0.890*cos33.6 = 0.741 m/s.
Yo = Ver. = 0.890*sin33.6 = 0.493 m/s.

Tr = (Yf-Yo) / g,
Tr = (0-0.493) / -9.8 = 0.0503 s. = Rise time or time to reach max. ht.

hmax = (Yf^2-Yo^2) / 2g,
hmax = (0-(0.493)^2) / -19.6=0.0124 m.

Tr + Tf = 0.0750 s.
0.0503 + Tf = 0.0750,
Tf = 0.0247 s. = Fall time.

d = Vo*t + 0.5g*t*t^2,
d = 0 + 4.9(0.0247)^2 = 0.0030 m.=Fall
distance.

hmax - d = 0.0124 - 0.003=9.4*10^-3 m
= 9.4 mm. = Thickness of mag.