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March 30, 2015

March 30, 2015

Posted by **FATHIMA** on Sunday, August 28, 2011 at 5:23am.

A: WHAT IS THA TYPE OF MOSSION MADE BY THE FROG?

B: FIND THE ANGLE THAT THE FGOGS INITIAL VELOCITY MAKES WITH THE HORIZONTAL,FIND THE MINIMUM DISTENCE THE INSECT HAVE 2 KEEP FROM THE FROG SO THAT IT CAN ESCAPE FROM THE FROG?

- +1 SCIENCE PHYSICS -
**Damon**, Sunday, August 28, 2011 at 6:49amThe path of the frog is parabolic

u = 20 cos A = constant

d = range = u t = 20 cos A t = 30

so t cos A = 1.5 and t = 1.5/cos A

vi = 20 sin A

v = 20 sin A - 9.8 t

h = 0 + 20 t sin A - 4.9 t^2

h = 0 when frog hits bug

0 = t (20 sin A - 4.9 t)

so t = 4.08 sin A

so

4.08 sin A = 1.5/cos A

and

sin A cos A = .3675

sin A (1-sin^2A)^.5 = .3675

(1-sin^2 A)^.5 = .3675/sin A

1 - sin^2 A = .135/sin^2 A

sin^2 A - sin^4 A = .135

sin^4 A - sin^2 A + .135 = 0

let x = sin^2 A

x^2 - x + .135 = 0

x = [ 1 +/-sqrt (1-.54) ]/2

x = [ 1 +/- .678 ]/ 2

x = sin^2 A = .322 or .839

sin A = .567 or .916

A = 34.5 deg or 66.3 deg

max range when A = 45 deg

u = 20 cos 45 = 14.1

vi = 20 sin 45 = 14.1

time in air

v = vi - g t

top when v = 0 halfway through flight

t at top = 14.1/9.8 = 1.44

so time in air = 2.88 seconds

d = u t = 14.1 * 2.88 = 40.7 meters

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