Biomechanics Quadratic Equation problem

[dsm63]

Homework Statement

1.
Neglecting the height of release, a ball is thrown vertically upwards at 20 m/s, find:
d) time (s) that the projectile is at height 10.5 m.

V1 = 20m/s
a = -9.81m/s^2
d = 10.5m

Homework Equations

d = v1*t + 1/2 a(t^2)

The Attempt at a Solution

subsitutiting i can get the formula to:

10.5 = 20t + 1/2(-9.81)t^2
10.5 -20t +4.905t^2 = 0
4.905t^2 - 20t = -10.5
t^2 - 20/4.905 t = -10.5/4.905

However I am rusty on my quadratic equation work and do not know how to solve from here and cannot seem to figure it out? any help would be awesome Thanks in advance.

 

[tiny-tim]

welcome to pf!

hi dsm63! welcome to pf!

(try using the X² button just above the Reply box )

10.5 -20t +4.905t² = 0

either use the formula [-b ± √(b2 - 4ac)]/2a,

or complete the square

 

[dsm63]

would i be right to assume that
for

[-b ± √(b2 - 4ac)]/2a

a = 4.905
b = -20
c = 10.5

when using this i get

t = 20 ±√19.7747

t = 24.44 or 15.55

this answer does not make sense? i think i have some something wrong..

using a calculator online i recieved

t = .619 or 3.45s

 

[tiny-tim]

(just got up :zzz:)

would i be right to assume that
for

[-b ± √(b² - 4ac)]/2a

that's correct

but you've done -b ± √[(b² - 4ac)/2a] !

 

[asteriatic]

I would first commend you for recognizing the need for precise equations and calculations in solving this problem. Your attempt at a solution is a good start, but as you mentioned, you are rusty on your quadratic equation work. Let me guide you through the steps to solve this problem.

First, let's rearrange the equation to the standard quadratic form: ax^2 + bx + c = 0, where a = 4.905, b = -20, and c = -10.5.

Next, we can use the quadratic formula to solve for t: t = (-b ± √(b^2 - 4ac)) / 2a. Plugging in the values, we get:

t = (-(-20) ± √((-20)^2 - 4(4.905)(-10.5))) / 2(4.905)
t = (20 ± √(400 + 205.8)) / 9.81
t = (20 ± √(605.8)) / 9.81
t = (20 ± 24.62) / 9.81

This gives us two possible solutions: t = 4.87 seconds or t = -0.43 seconds. Since time cannot be negative in this case, we can discard the negative solution and conclude that the projectile is at a height of 10.5 m after 4.87 seconds.

I hope this helps you better understand how to solve quadratic equations. Keep practicing and you'll get better at it!