Calculate Speed of Red Ball Falling From 26m

[Thewindyfan]

Homework Statement

A red ball is thrown down with an initial speed of 1.3 m/s from a height of 26 meters above the ground. The force of gravity due to the Earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

What is the speed of the red ball right before it hits the ground?

Homework Equations

x = 26 - 1.3t - 4.905t^2

The Attempt at a Solution

I get how to do this problem, but I wasn't able to get the correct answer the first time when I did it completely mathematically. Can someone check where I'm going wrong in this work?

We need to find the time that the ball hits the ground --> set x = 0
0 = 26 - 1.3t - 4.905t^2
-26 = -1.3t - 4.905t^2
-26 = (-1.3 - 4.905t)*t
t = 0,
-4.905t = -24.7
t = 5.03 seconds

I don't know where I'm going wrong to get that t = 5.03 seconds, when the answer is 2.17 seconds (by using a calculator and plotting the zeroes). Is it wrong to solve the quadratic like this when solving for the roots?

 

[Ray Vickson]

Homework Statement

A red ball is thrown down with an initial speed of 1.3 m/s from a height of 26 meters above the ground. The force of gravity due to the Earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

What is the speed of the red ball right before it hits the ground?

Homework Equations

x = 26 - 1.3t - 4.905t^2

The Attempt at a Solution

I get how to do this problem, but I wasn't able to get the correct answer the first time when I did it completely mathematically. Can someone check where I'm going wrong in this work?

We need to find the time that the ball hits the ground --> set x = 0
0 = 26 - 1.3t - 4.905t^2
-26 = -1.3t - 4.905t^2
-26 = (-1.3 - 4.905t)*t
t = 0,
-4.905t = -24.7
t = 5.03 seconds

I don't know where I'm going wrong to get that t = 5.03 seconds, when the answer is 2.17 seconds (by using a calculator and plotting the zeroes). Is it wrong to solve the quadratic like this when solving for the roots?

If $-26 = t(-1.3-4.905 t)$ you cannot have $t = 0$. If you set $t = 0$ you get right-hand-side = 0 (because 0 times anything = 0), but left-hand-side = -26, and 26 ≠ 0.
So, yes, it is very wrong indeed to solve quadratic equations that way!

Why don't you just use the familiar quadratic equation roots formula?

 

[Thewindyfan]

If $-26 = t(-1.3-4.905 t)$ you cannot have $t = 0$. If you set $t = 0$ you get right-hand-side = 0 (because 0 times anything = 0), but left-hand-side = -26, and 26 ≠ 0.
So, yes, it is very wrong indeed to solve quadratic equations that way!

Why don't you just use the familiar quadratic equation roots formula?

Yeah I guess I just confused that approach with solving the time for when x = some number. Thanks for pointing that out!