TBR Physics Translational Motion

[Ihavesomanyquestions]

Approximately how high above the ground is a spherical projectile nine seconds after it has been launched straight up with an initial velocity of 60 m/s?

A. 11.1 m
B. 44.4 m
C. 138.2
D 320.7

In the answer explanation they find the time it takes for the projectile to reach its apex, which is six seconds, for a total flight time of 12 seconds. From their they conclude that the height at 9 seconds is equal to the height at 3 seconds. How did they figure that out based on this information? I know it is used to make the calculation easier. Either using t=3 or t=9 you get the same answer of 135. If this were on the MCAT I would have thought that I did the problem wrong because the calculation is not among the answer choice.

The first time I did this calculation I got the answer wrong because I had the equation as y=vt + 1/2at^2 instead of y=vt - 1/2at^2. Can someone explain the negative sign to me? I know that it is due to gravity.

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[JMMTB]

Velocity and acceleration are vectors. The acceleration points straight down and the velocity is straight up (initially) because it initially received a large force upward (force is gone once released).

v=at for free falling objects, but it was launched upward with a velocity of 60m/s so now we have v=vo-at (minus because opposite directions) and we have decided that the ground is the origin. Everything above it is positive and below negative.
It reaches the apex when v=0 and we can solve 0=60-at
60=10t
t=6 seconds to reach the apex

H is the same for t=3 or 9 because it is parabolic. From the apex we have symmetry so we can conclude that h is the same for t=3 or 9

y=-.5at^2+vot
180-5*3^2=135

Good luck studying!