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?
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: The type of motion made by the frog can be classified as projectile motion. When the frog jumps, it follows a curved path due to the combination of its initial horizontal velocity and vertical acceleration due to gravity.
B: To find the angle that the frog's initial velocity makes with the horizontal, we can use trigonometry. Let's assume the angle made by the frog's initial velocity with the horizontal is θ.
Using the formula for horizontal distance in projectile motion, we have:
horizontal distance = initial velocity * time * cosine(θ)
In this case, the horizontal distance is given as 30 meters and the initial velocity is given as 20 meters/second. We know that the frog lands on the insect, so the time taken to reach the insect can be calculated using the distance formula:
time = horizontal distance / (initial velocity * cosine(θ))
Substituting the given values, we have:
30 = 20 * time * cosine(θ)
Simplifying the equation, we get:
cosine(θ) = 1.5 / time
To find the minimum distance the insect should keep from the frog to escape, we need to determine the maximum range of the frog's jump. The range is the horizontal distance traveled by the frog when it lands on the insect.
The formula for range in projectile motion is:
range = initial velocity^2 * sine(2θ) / gravitational acceleration
In this case, the gravitational acceleration can be approximated as 9.8 meters/second^2.
Substituting the values, we have:
range = (20^2 * sine(2θ)) / 9.8
To find the minimum distance the insect should keep, it should be greater than the range.
Therefore, the minimum distance the insect should keep from the frog to escape is greater than the calculated range.