Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.
Therefore,
u = 10 m/s, initial upward velocity.
H = - 20 m, position of the ground.
g = 9.8 m/s², acceleration due to gravity.
Part (a)
When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore
u² - 2gh = 0
h = u²/(2g) = 10²/(2*9.8) = 5.102 m
At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.
Answer: 25.1 m above ground
Part (b)
Let v = the velocity when the frog hits the ground. Then
v² = u² - 2gH
v² = 10² - 2*9.8*(-20) = 492
v = 22.18 m/s
Answer: The frog hits the ground with a velocity of 22.2 m/s
If you are talking about sound frequency you need to consider what area you are in because in a concert hall it is big and helps the sound spread but in an airplane it is the opposite.
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Answer:
![\theta=50.79^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D50.79%5E%7B%5Ccirc%7D)
Explanation:
The radius of curvature of the velodrome, r = 20 m
Speed of cyclists, v = 15.5 m/s
Let
is the angle at which the surface should be banked. The horizontal and vertical forces acting on the cyclists are as follows :
![F_x=\dfrac{mv^2}{r}](https://tex.z-dn.net/?f=F_x%3D%5Cdfrac%7Bmv%5E2%7D%7Br%7D)
![F_y=mg](https://tex.z-dn.net/?f=F_y%3Dmg)
From the above two equations,
![tan\theta=\dfrac{v^2}{rg}](https://tex.z-dn.net/?f=tan%5Ctheta%3D%5Cdfrac%7Bv%5E2%7D%7Brg%7D)
![tan\theta=\dfrac{(15.5)^2}{20\times 9.8}](https://tex.z-dn.net/?f=tan%5Ctheta%3D%5Cdfrac%7B%2815.5%29%5E2%7D%7B20%5Ctimes%209.8%7D)
![tan\theta=1.225](https://tex.z-dn.net/?f=tan%5Ctheta%3D1.225)
![\theta=50.79^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D50.79%5E%7B%5Ccirc%7D)
So, the angle at which the surface is banked is 50.79 degrees. Hence, this is the required solution.