because this time t is a known quantity. The answer is (B) 4 m/s^2 , less than half of Earth’s
gravitational field, but close to Mars’s gravitational field.3 . First find the velocity function by taking the derivative of the position function: v (t ) = 6t − 4. Now
plug in t = 6 to get the velocity after 6 s; you get 32 m/s. Note that this velocity is positive. Was the
object originally moving in the positive direction? Plug in t = 0 to the velocity formula to find out ...
you find the initial velocity to be −4 m/s, so the object was originally moving in the negative direction,
and has reversed direction after 6 s. The answer is (E).
4 . (a) Because the angle 30° is measured to the horizontal, the magnitude of the vertical component of the
velocity vector is just (200 m/s) (sin 30°), which is 100 m/s. The direction is “up,” because the plane
is flying up.
(b) The horizontal velocity of a projectile is constant. Thus, the horizontal velocity when the bomb hits
the target is the same as the horizontal velocity at release, or (200 m/s)(cos 30°) = 170 m/s, to the
right.
(c) Let’s call “up” the positive direction. We can solve this projectile motion problem by our table
method.
Don’t forget to convert to meters, and be careful about directions in the vertical chart.
The horizontal chart cannot be solved for time; however, the vertical chart can. Though you
could use the quadratic formula or your fancy calculator to solve x − x 0 = v 0 t + ^1 / 2 at 2 , it’smuch easier to start with **, v (^) f 2 = v 0 2 + 2a (x − x 0 ), to find that vf vertically is −220 m/s (this
velocity must have a negative sign because the bomb is moving down when it hits the ground).
Then, plug in to (v (^) f = v 0 + at ) to find that the bomb took 32 s to hit the ground.
(d) Start by finding how far the bomb went horizontally. Because horizontal velocity is constant, we
can say distance = velocity × time. Plugging in values from the table, distance = (170 m/s)(32 s) =
5400 m. Okay, now look at a triangle: