What are the times when a pellet reaches 25 meters above its initial height?

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Homework Help Overview

The problem involves a pellet fired upward with an initial velocity of 25 m/s, and the goal is to determine the times when it reaches a height of 25 meters above its initial position. The context is within kinematics, specifically projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set up the equation of motion but becomes confused after simplifying it. They express uncertainty about using the quadratic formula. Other participants question the restrictions on using the quadratic formula and suggest alternative methods such as finding final velocity or completing the square.

Discussion Status

Participants are exploring different methods to solve the equation, with some suggesting the use of the quadratic formula while others provide alternative approaches. There is an ongoing dialogue about the appropriateness of various mathematical techniques given the original poster's educational background.

Contextual Notes

The original poster indicates that they are in grade 10 and have not yet learned the quadratic formula, which influences the discussion on how to approach the problem.

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Homework Statement


A pellet is fired upward with an initial velocity of 25 m/s. At what times in the pellet 25 meters above its fired height


Homework Equations


S=ut+1/2at^2 I got it down to,
Displacement(s)=25m
U(initial velocity)=25 m/s
and a=acceleration due to gravity -9.81
I got 25=25t+1/2(-9.81)(t^2)
then 25=25t+-4.91t^2 and I am stumped

The Attempt at a Solution



I tried dividing both sides by 25t, leaving
1/t=-4.91t^2, but I really have no clue.
I can't use the quadratic formula and I need 2 times. Any hints would be very nice! Thank you.
 
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I'm a little bit confused, are you not allowed to use the quadratic formula for this problem? The quadratic formula is the perfect way to solve for those two times.
 
kiwikahuna said:
I'm a little bit confused, are you not allowed to use the quadratic formula for this problem? The quadratic formula is the perfect way to solve for those two times.
Well I am in grade 10, and we haven't learned it yet, so it is off limits. Sorry for the confusion
 
In that case, you should find the final velocity of this projectile motion. After you find the final velocity, you can find the times using the definition of acceleration.
 
Well, let's start with what you arrived at:

25=25t+-4.91t^2

Now just factor this equation to find the values of t. You probably know the method of completing the squares, that is :

-4.9t^2 + 25t = 25

Dividing through whatever is multiplied on the square term

-t^2 + (25/4.9)t = 25/4.9

making the squared term positive

t^2 - (25/4.9)t = -25/4.9


Half the x coefficient (-25/4.9) -> (-25/9.8) and squaring it: (625/96.04) then adding this to both sides:

t^2 - (25/4.9)t + 625/96.04 = -25/4.9 + 625/96.04

then you know:

(t - 25/9.8)^2 = (625(4.9) - 25(96.04))/(4.9)(96.04)

simplifying:

(t-25/9.8)^2 = 661.5/470.596

Thus you know that

t - 25/9.8 = +/- sqrt (661.5/470.596)

as both the negative and positive numbers make a square positive number.

and thus your answers:

t = 25/9.8 +/- sqrt (661.5/470.596)

and using your trusty calculator you know now :

t = 3.736627555 (for the +)
and
t = 1.365413261 (for the -)
 

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