Hawk velocity/angle problem

[Melchior25]

Homework Statement

A hawk is flying horizontally at 14.0 m/s in a straight line, 250 m above the ground. A mouse it has been carrying struggles free from its grasp. The hawk continues on its path at the same speed for 2.00 s before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse 3.00 m above the ground.

(a) Assuming no air resistance, find the diving speed of the hawk.
m/s
(b) What angle did the hawk make with the horizontal during its descent?
° (below the horizontal)
(c) For how long did the mouse "enjoy" free fall?
s

Homework Equations

h=(.5)gt^2
vf=vi+at

xf=vxi*t

The Attempt at a Solution

for (a) I used h=(.5)gt^2 and solved for t. t=7.09987 s
I then plugged t into vf=vi+at to get a final v of 83.6497m/s diving speed for the hawk. Does this look right?

For (b) I'm lost on where to start. I drew a triangle and have a y height of 247. I also have a 90 degree angle. I don't know however how to solve for the angle.

For (c) I was going to use xf=vxi*t. Knowing my xf already and initial velocity once I figure out the angle from part (b).

Please help. I'm so lost.

 

[Melchior25]

So I figured out part a and part c but I'm struggling with art b. A is 50.0297 m/s and the time the hawk dove was 5.09987 s. I multiplied these two and got my hyp. I also know my y length. I drew a triangle and thought I calculated theta but my answer is wrong.

 

[Moetasim]

Your approach for part (a) is correct. However, there is a mistake in your calculation for t. The correct value for t is 9.8995 s. Using this value, the diving speed of the hawk is 138.659 m/s.

For part (b), we can use the law of cosines to find the angle. The law of cosines states that c^2 = a^2 + b^2 - 2abcosC, where c is the side opposite angle C. In this case, c is the diving speed of the hawk, a is the horizontal velocity of the hawk (14 m/s), and b is the vertical distance traveled by the hawk (250 m - 3 m = 247 m). Solving for C, we get an angle of 81.47 degrees below the horizontal.

For part (c), you can use the equation h=(.5)gt^2 to find the time of free fall for the mouse. Since the initial height is 250 m and the final height is 3 m, we can set up the equation as 3 = (.5)(9.8)t^2. Solving for t, we get a time of 2.18 seconds for the mouse's free fall.

Overall, your approach was on the right track. Just be sure to double check your calculations and use the appropriate equations.