Question

An eagle is flying horizontally at 5.2 m/s with a fish in its claws. It accidentally drops the fish. (a) How much time passes before the fish's speed quadruples? (b) How much additional time would be required for the fish's speed to quadruple again?

Answer 1

a)

Along the horizontal direction , the velocity remains constant as 5.2 m/s.

v_ox= initial speed in x-direction = 5.2 m/s

v_oy = initial speed in y-direction = 0 m/s

so initial speed is given as

v_o = sqrt(v_ox² + v_oy²) = sqrt(5.2² + 0²) = 5.2 m/s

Given that : v_f = final speed = 4 v_o = 4 (5.2) = 20.8 m/s

sqrt(v_fx² + v_fy²) = 20.8

sqrt(5.2² + v_fy²) = 20.8

v_fy = 20.14 m/s

a_y= acceleration along the vertical direction = 9.8 m/s²

t = time of travel

using the equation

v_fy = v_oy + a_y t

20.14 = 0 + (9.8) t

t = 2.1 sec

b)

Along the horizontal direction , the velocity remains constant as 5.2 m/s.

v_ox= initial speed in x-direction = 5.2 m/s

v_oy = initial speed in y-direction = 20.14 m/s

so initial speed is given as

v_o = sqrt(v_ox² + v_oy²) = sqrt(5.2² + 20.14²) = 20.8 m/s

Given that : v_f = final speed = 4 v_o = 4 (20.8) = 83.2 m/s

sqrt(v_fx² + v_fy²) = 83.2

sqrt(5.2² + v_fy²) = 83.2

v_fy = 83.04 m/s

a_y= acceleration along the vertical direction = 9.8 m/s²

t = time of travel

using the equation

v_fy = v_oy + a_y t

83.04 = 0 + (9.8) t

t = 8.5 sec