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?
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?