If a 1500 kg projectile is dropped from a height of 50 m, with what approximate speed will the projectile strike the ground?
A) 10 m/s
B) 30 m/s
C) 75 m/s
D) 100 m/s
Click for Explanation
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.
Want more MCAT practice?
We’ve got options for every schedule and learning style!
Whether you need the flexibility of a Self-Paced Course, the instruction of a live 515+ Course, or the 1:1 attention of a private MCAT tutor, Blueprint MCAT has the MCAT prep option that works for your learning style!
Get started with a free MCAT diagnostic, one free practice exam, and tons more MCAT prep resources.
Not sure which option is right for you? Schedule a free MCAT consultation with an MCAT expert using the form below. No obligation, just expert advice.
