Physics, 29.11.2019 00:31, arivalen

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. unfortunately the archer stands on an elevated platform of unknown height. however, you find the arrow stuck in the ground 67.0 m away, making a 3.00 ∘ angle with the ground. how fast was the arrow shot?

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To develop this problem it is necessary to apply the kinematic equations that describe displacement, velocity and clarification.

By definition we know that velocity is defined as the change of position due to time, therefore

$V = \frac{d}{t}$

Where,

d = Distance

t = Time

Speed can also be expressed in vector form through its components V_x and V_y

In the case of the horizontal component X, we have to

$V_x = \frac{d}{t}$

Here d means the horizontal displacement, then

$t = \frac{d}{V_x}$

$t = \frac{67}{V_x}$

At the same time we have that the vertical component of the velocity is

V_y = gt

Here,

g = Gravity

Therefore using the relation previously found we have that

$V_y = g \frac{67}{V_x}$

The relationship between the two velocities and the angle can be expressed through the Tangent, therefore

$tan\theta = \frac{V_y}{V_x}$

$tan \theta = \frac{g \frac{67}{V_x} }{V_x}$

$tan 3 = \frac{9.8\frac{67}{V_x} }{V_x}$

$tan 3 = \frac{9.8*67}{V_x^2}$

$V_x^2 = \frac{9.8*67}{tan 3}$

$V_x= \sqrt{ \frac{9.8*67}{tan 3}}$

$V_x = 111.93m/s \hat{i}$

This is the horizontal component, we could also find the vertical speed and the value of the total speed with the information given,

Then V_y,

$V_y = g \frac{67}{V_x}$

$V_y = 9.8*\frac{67}{111.93}$

$V_y = 5.866m/s\hat{j}$

$|\vec{V}| = \sqrt{111.93^2+5.866^2}$

$|\vec{V}| = 112.084m/s$

answered: Guest

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answered: Guest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.