MCAT Physics Question — Dropped Projectile
- Oct 27, 2014
- MCAT Question of the Day
A) 10 m/s
B) 30 m/s
C) 75 m/s
D) 100 m/s
This question is testing your understanding of projectile motion and application of kinematic equations. The question stem states that the projectile is dropped so you can assume that the initial velocity, vo, is equal to zero. Setting y = 0 to be at the surface, the initial height is 50 m and the final height is 0 m, so the change in height, Δh, is equal to -50m. Lastly the projectile accelerates at a rate of gravity (-10 m/s assuming the positive direction is upwards). Since we know vo, Δh, and a, and the question asks for the final velocity, vf, you can use the equation, vf2 = vo2 + 2aΔx.
The equation simplifies to the following:
vf = √2aΔx = √2(−10 m/s)(−50 m) = ,-1000.=33.33 m/s
Thus, B is the correct answer.
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