The 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
A FROG SAW AN INSECT ON THE GROUND AT A HORIZONTAL DISTANCE OF 30 METER FROM IT. THE FROG CAN JUMP WITH A SPEED 20 METER/SECOND IN ANY DIRECTION. IT JUMPS WITH HIS INITIAL SPEED AND LAND ON THE INSECT.
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 answer
3 answers