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stepan [7]
3 years ago
15

A cannonball is fired horizontally from the top of a cliff. the cannon is at height h = 80.0 m above ground level, and the ball

is fired with initial horizontal speed v0. assume acceleration due to gravity to be g = 9.80 m/s2 .assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. what is the y position of the cannonball at the time tg/2?
Physics
1 answer:
SOVA2 [1]3 years ago
6 0

Answer;

= 60 m

Explanation;

s = 80 m

= ½at² = ½ * 9.8m/s² * t² → t = 4.04 s  

At any time 0 ≤ t ≤ 4.04 s, height is  

y = Yo + ½at² = 80 m - 4.9m/s² * t²  

so at t = 4.04s/2 = 2.02 s,  

y = 80 m - 4.9m/s² * (2.02s)² = 60 m  

(so it drops 15 m during the first 2.02 s and 60 m during the second 2.02 s)

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Answer:

E = (-3.61^i+1.02^j) N/C

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Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

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q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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2 years ago
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so the correct answer is D option

7 0
3 years ago
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