A hawk is flying horizontally at 17.0 m/s in a straight line, 160. m above the g

Question

A hawk is flying horizontally at 17.0 m/s in a straight line, 160. 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 2.00s 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. Hint: From the hawk's point of view, what is the motion of the mouse? Does the hawk have to change both components of its motion to catch the mouse? Do part (c) first, before the other parts (a) Assuming no air resistance, find the diving speed (magnitude of the total velocity vector) of the hawk. (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?

Explanation / Answer

a)

from

Y=Yo+Vot+(1/2)at2

since it hits ground finally ,y=0

=>0=160+0-(1/2)*9.8*t2

t=5.71 s

The raptor dives for

t'=5.71-2=3.71 s undergoing displacement 160 m downward and 17*3.71=63.14 m forward

therefore

V=d/t'=sqrt(1602+63.142)/3.71

V=46.3 m/s

b)

angle

o=tan-1(-160/63.14)=-68.46o

c)

time taken by the mouse during freefall is

t=5.71 s

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