CATAGORY: Mechanics QUESTION: Ok, I'm studying physics over the summer on my own, and I bought the text book, but sometimes the answers in the book don't seem reasonable. I'd like a 2nd opinion. The question is: "Agent Tim, flying a constant 185 Km/h horizontally in a low-flying helicopter, wants to drop a small explosive onto a master criminal's automobile traveling 145 Km/h (same direction) on a level highway 88.0 meters below. At what ANGLE (with the horizontal) should the car be in his sights when the bomb is released?" By using the formula: [ TAN(-1)angle = -9.80 t / horizontal velocity ] ........ i get -75 degrees. Answer in book is -61.8 degrees am i wrong, or is the book wrong?

Back to catalog.

Since the boms horizontal and vertical motions aer decoupled we can
consider first how long it takes the bomb to fall.  Since it has no
initial vertical velocity the equation of motion is

(1)   d = do -.5gt^2

where do = 88.0m, g=9.8m/s^2 t is time and d = 0 (when the bomb hits
the car) Thus the time it takes to fall is (solve (1) for t with d=0m)

(2)  t = sqrt(do/(.5g)) = sqrt(88.0/(.5x9.8m/s^2)) = 4.23s

The horizontal motion of the bomb is not affected by gravity.  It is
released at 185 km/hr so it continues to move at 185km/hr.  The
velocity of the bomb relative to the car however is 185 km/hr - 145
km/hr = 40 km/hr or 40km/hr x 1hr/3600s x 1000m/1km = 11.1 m/s.  This
is how fast the bomb is approaching the car (horizontaly).  In 4.23s
the distance traveled is

(3)  X = 4.23s X 11.1m/s = 47 m

>From the figure below we see that

(4)        tan(a) = 88m/47m = 1.872

so

(5)            a = 61.8 below horizontal

_________^_________
X        __|_
\______/   0\__
\______/  --> 185 km/hr
__/___/___,
|\) a
| \
|  \
|   \
88.0m |    \
|     \
|  47m \
|_______\__
___/| | |\___
`\__^____^___>'--> 145 km/hr
_______________________________O____O__________________