Physics, 14.07.2021 14:00 fahad7007

When solving projectiles you will often get two possible times for solutions. Sometimes a time will be negative and can be rejected. Other times both times will be positive and therefore possible. Explain why both times above make sense in the context of the given information used. Which time will you ultimately choose and why? Context is attached below.

When solving projectiles you will often get two possible times for solutions. Sometimes a time will

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Answer from: Lacrosse34

The reason why times, 0.23 s, and 1.80 s makes sense is that a projectile passes through a given height level, which is lesser than the maximum height reached by the projectile, twice in its trajectory

Explanation:

The formula for the time of motion of a projectile is given as follows;

$t = \dfrac{-V_{oy} \pm \sqrt{V_{oy} ^2 - 2 \cdot g \cdot \Delta y} }{-g}$

Therefore, when $V_{oy} ^2 = 2 \cdot g \cdot \Delta y}$ , we have only one time value

When $V_{oy} ^2 2 \cdot g \cdot \Delta y}$ , two time values can be obtained and both will be positive when we have;

$V_{oy} \sqrt{V_{oy} ^2 - 2 \cdot g \cdot \Delta y} }$

When $V_{oy} < \sqrt{V_{oy} ^2 - 2 \cdot g \cdot \Delta y} }$ , one of the time values will be negative and can be rejected

Therefore, given that the times obtained above are 1.80 s, and 0.23 s, and both make sense due to the following reason;

In the path of the projectile motion of the basketball, there are two points in time at which the height of the basketball above the starting point is exactly 2 meters, given that the maximum height reached is more than 2 meters

The first time the basketball is 2 meters above the point it is shot is lesser of the two calculated time values, which is during the upward motion of the basketball before it reaches the maximum height, while the second time is, which is the larger calculated time, is the time that the basketball reaches the hoop, after flying past the highest point

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

Answer from: Quest

b.a device that inputs heat into a cool object

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