Question

Chapter 03, Problem 17 A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.840 m/s at an angle of 39.0° above the table, and it lands on the magazine 0.0650s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters. Number Units the tolerance is +/-2%

Answer 1

Solution)

Let thickness of the magazine be y

The spider lands on the magazine 0.0650 secs later

Y component of acceleration (Ay)= -9.8m/s2

Initial velocity (v0)=+0.840 m/s

Given,Direction of initial velocity (θ)=39 degrees

Time (t)=0.0650sec

Voy=vo*sin 39 =0.840*sin 39=0.528 m/s

Now, we know

S=ut+0.5at^2

S=0.528*0.0650 + 0.5*(-9.8)*(0.0650)^2=0.0136 m=13.65 mm (Ans)

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