Physics, 07.07.2019 16:30 steven123pink

An archer shoots an arrow horizontally that hits a target 100m away. a bystander times the flight of the arrow at 2.2s. a) what was the initial velocity of the arrow? b) how far did the arrow drop in the vertical direction?

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Answer from: AlaishaWiseBrown

The archer is fired horizontally.

It hits the target which is at 100 m distance.

The horizontal velocity of the archer is always constant for a projectile in this situation.

The total time of flight is given as 2.2 s

Let the initial velocity is u .

Hence the target distance = ut

⇒ 100 m = u×2.2 s

⇒u = 100÷2.2 m/s

=45.45 m/s

We are asked to calculate the vertical distance travelled.

The vertical distance travelled is calculated as-

$s= ut +\frac{1}{2} at^2$ [ s is the distance and a is the acceleration]

$s = 0*2.2 -\frac{1}{2} g[2.2]^2$ [g is te acceleration due to gravity]

$s=\frac{1}{2} 9.8*[2.2]^2$ [here we have taken only magnitude]

=23.716 m [ans]

Answer from: Quest

carla slowed down in the beginning of her trip, stopped, and then sped up

Answer from: Quest

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.

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

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

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