Crickets chirpy and milada jump from the top of a vertical cliff. chirpy drops downward and reaches the ground in 2.70 s, while
milada jumps horizontally with an initial speed of 95.0 cm>s. how far from the base of the cliff will milada hit the ground? ignore air resistance.
1 answer:
The individual directions of motion of Chirpy and Milada are components of projectile motion. A projectile motion is characterized by a motion in the shape of an arc. The thing about projectile motion is, the horizontal component and the vertical component are independent of each other. The horizontal motion acts on constant velocity, while the vertical motion acts on a constant acceleration equal to the force of gravity, 9.81 m/s². Even though they are independent, there is a relationship between them called the trajectory of a projectile equation:
y = xtanθ + gx²/2v²cos²θ
where
y is the vertical height
x is the horizontal range
θ is the angle of inclination
g is 9.81 m/s²
v is the initial velocity
Since Milada jumps horizontally, there is no angle of inclination: θ=0°. The initial velocity is equal to 95 cm/s or 0.95 m/s. Now, we have to determine x. But we can't do that without finding y first. This can be obtained from Chirpy's downward motion. In a free falling motion, the time of flight is equal to
t = √2y/g
2.70 = √2y/9.81
y = 35.757 m
Now, we can solve for x. I suggest you use your scientific calculator so that you can easily solve for x.
35.757 = xtan0° + (9.81)x²/2(0.95)²cos²0°
x = 2.565
Therefore, Melinda hits the ground 2.565 meters away from the base of the cliff.
y = xtanθ + gx²/2v²cos²θ
where
y is the vertical height
x is the horizontal range
θ is the angle of inclination
g is 9.81 m/s²
v is the initial velocity
Since Milada jumps horizontally, there is no angle of inclination: θ=0°. The initial velocity is equal to 95 cm/s or 0.95 m/s. Now, we have to determine x. But we can't do that without finding y first. This can be obtained from Chirpy's downward motion. In a free falling motion, the time of flight is equal to
t = √2y/g
2.70 = √2y/9.81
y = 35.757 m
Now, we can solve for x. I suggest you use your scientific calculator so that you can easily solve for x.
35.757 = xtan0° + (9.81)x²/2(0.95)²cos²0°
x = 2.565
Therefore, Melinda hits the ground 2.565 meters away from the base of the cliff.
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Answer:
For the point inside the cylinder:
For the point outside the cylinder:
where x0 is the position of the point on the x-axis and σ is the surface charge density.
Explanation:
Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.
<u>x = -x0:</u>
The cylinder is consist of the sum of the rings with the same radius.
First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.
We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.
The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.
Now, the electric field created by the small portion is 'dE'.
The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.
We have to integrate it over the ring, which is an angular integration.
This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:
The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.
<u>x = +x0: </u>
We will only change the boundaries of the last integration.