Kevin S. Huang

2012: Problems 6-7

Topic: Kinematics
Concepts: Fictitious forces, Free fall

Solution:

We can go to the free fall frame accelerating downwards at $g$ so that the projectiles become free particles (since we include the fictitious inertial force of magnitude $mg$ pointing upwards which cancels out gravity). Then we have

h=(v_1+v_2)t

t=\frac{h}{v_1+v_2}=\frac{200\,\mathrm{m}}{25\,\mathrm{m/s}+55\,\mathrm{m/s}}=2.5\,\mathrm{s}

so the answer is B.

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Topic: Kinematics
Concepts: Free fall

Solution:

Since we found in the previous problem that the projectiles collide after time $t=2.5\,\mathrm{s}$ , the top projectile moves down by a distance

d=v_2t+\frac{1}{2}gt^2=(55\,\mathrm{m/s})(2.5\,\mathrm{s})+\frac{1}{2}(10\,\mathrm{m/s^2})(2.5\,\mathrm{s})^2=170\,\mathrm{m}

so the answer is E.

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Kevin S. Huang

2012: Problems 6-7

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2012: Problems 6-7

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