#2 When the Sun is directly overhead, a hawk dives toward the ground with a cons

Question

#2

When the Sun is directly overhead, a hawk dives toward the ground with a constant velocity of 4.50 m/s at 60.0° below the horizontal. Calculate the speed of its shadow on the level ground m/s Neud Help? Suppose that the position vector for a particle is given as a function of time by:(t)-x(t)I +y(t)J, with x(t) = at + b and y(t) = ct2 + d, where a = 2.00 m/s, b = 1.10 m, c = 0.11 8 m/s2, and d = 1 .08 m. (a) Calculate the average velocity during the time interval from t = 1 .85 s to t = 4.10 s. (b) Determine the velocity at t = 1 .85 s Determine the speed at t 1.85 s m/s Need Hlp? A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by , and .454 4AF Where and y are in meters and t is in seconds (a) Write a vector expression for the ball's position as a function of time, using the unit vectors i andj (Give the answer in terms of t.)

Explanation / Answer

The hawk is moving at 60 deg below the hroizontal. It valicty can be resolved along the verticle and horizontal.

If the hawk was just descending vertically without any horizontal movemnt then its shadow remains at the same pos.

=> the vertical component of the velocity has no impact on shadow movemnt.

only the horizontal component causes the shaodw to move.

velocity of the shadow = horizontal comp = 4.5*Cos(60) = 2.25 m/s

x(t) = at +b = 2t +1.1 m

y(t) = ct2 +d = 0.118t2 +1.08 m

at t= 1.85    x= 4.8 m/s     y = 1.48 m

at t= 4.1 ; x = 9.3 ; y = 3.06 m

displacement = 4.77 m, average vel = 4.77/2.25 =2.12 m/s

velcity at any instatan of time t is givne by

v(t) = dx/dt i + dy/dt j   = 2 i + 0.236t j

velocity at 1.85 v(1.85) = 2 i + 0.44 j

speed at t = 1.85   = sqr( 22 + 0.442) = 2.047 m/s

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