OK, so first I'd do the equation in the Y direction to find the answer to part 3).
X = Xo + Vot + 1/2at^2
3 = 180 + (0*t) + (1/2 * -9.8 * t^2)
3 = 180 - 4.9t^2
4.9 t^2 = 177
t^2 = 36.12
t = 6.01s
Now we go back to part 1). We know the Hawk waited 2s before starting its dive so it dove for 6.01 - 2 = 4.01 s. We also know that during that time, it traveled 180 - 3 = 177 m vertically. But, since there's no air resistance, the mouse kept traveling horizontally too! And, during those 6.01s, it traveled 10 * 6.01 = 60.1 m forward. Since the Hawk traveled 20 m in the 2 s it waited, it must have traveled 60.1 - 20 = 40.1 m horizontally during it's dive!
OK, so it dove 177 m vertically while traveling 40.1 m horizontally. Pythagorean theorem then says d = sqrt(177^2 + 40.1^2) = 181.5 m. Since it did that in 4.01 s, its speed must have been 181.5 / 4.01 = 45.26 m/s.
We can calculate the angle... Draw a picture and you may see...
tan(theta) = 177 / 40.1 = 4.414
theta = 1.348 rad or 77.2°
A hawk is flying horizontally at 10 m/s in a straight line 200m above the ground. A mouse it was carrying is released from its grasp. The hawk continues on its path at the same speed for two seconds before attempting to retrieve its prey. To accomplis the retrieval, it dives in a straight line at constant speed and recaptures the mouse 3m above the ground. Assuming no air resistance
a) find the diving speed of the hawk
b)what angle did the hawk make with the horizontal during its descent?
c)for how long did the mouse enjoy free fall?
1 answer