<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal
it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
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let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
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t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s
so it will able to cross</span>
Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for DAC:
⇒
Now for the BAC:
⇒
Now, differentiating w.r.t x:
For maximum angle, = 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 =
After solving the above eqn, we get
x =
The observer should stand at a distance equal to x =
I think it would be constellations
If that happens to your car and you double the speed then it will take the car longer to come to a complete stop
Answer: μ = 0.8885
Explanation: Force due to friction is calculated as:
At an inclined plane, normal force (N) is: N = mgcosθ, in which θ=32.51.
Power associated with work done by friction is . The variable x is displacement the object "spent its energy".
Power associated with work done by gravitational force is P = mghcosθ, where h is height.
The decline forms with horizontal plane a triangle as draw in the picture.
To determine force due to friction:
Replacing force:
Calculating h using trigonometric relations:
h = sin(32.51)
Coefficient of Kinetic friction is
μ = 0.8885
<u>For these conditions, coefficient of kinetic friction is </u><u>μ = 0.8885</u><u>.</u>