SOLUTION: Please provide detailed steps to solving the following word problem: A ball is thrown upward with an intial velocity of 14 meters per second (mps) from a cliff that is 70 meters

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Question 471488: Please provide detailed steps to solving the following word problem:
A ball is thrown upward with an intial velocity of 14 meters per second (mps) from a cliff that is 70 meters high. The height of the ball is given by the quadratic equation h = -4.9t^2 + 14t + 90 where h is in meters and t is the time in seconds since the ball was thrown. Find the time that the ball will be 20 meters from the ground...rounded to the nearest tenth of a second
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A ball is thrown upward with an initial velocity of 14 meters per second (mps)
from a cliff that is 70 meters high.
The height of the ball is given by the quadratic equation h = -4.9t^2 + 14t + 90
where h is in meters and t is the time in seconds since the ball was thrown.
Find the time that the ball will be 20 meters from the ground...rounded to the nearest tenth of a second
:
Just take the given equation, replace h with 20 and you have:
-4.9t^2 + 14t + 90 = 20
arrange it into a quadratic equation
-4.9t^2 + 14t + 90 - 20 = 0
-4.9t^2 + 14t + 70 = 0
Use the quadratic formula to find t
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this equation x=t; a=-4.9; b=14; c=70
t+=+%28-14+%2B-+sqrt%2814%5E2-4%2A-4.9%2A70+%29%29%2F%282%2A-4.9%29+
:
t+=+%28-14+%2B-+sqrt%28196-%28-1372%29+%29%29%2F%28-9.8%29+
:
t+=+%28-14+%2B-+sqrt%28196%2B1372+%29%29%2F%28-9.8%29+
:
t+=+%28-14+%2B-+sqrt%281568+%29%29%2F%28-9.8%29+
Two solutions
t+=+%28-14+%2B+39.6%29%2F%28-9.8%29+
t = 25.6%2F%28-9.8%29
t = -2.6, obviously this is not the solution
and
t+=+%28-14+-+39.6%29%2F%28-9.8%29+
t = %28-53.6%29%2F%28-9.8%29
t = 5.47 ~ 5.5 seconds it will be at 20 meter
:
:
You can prove this to yourself, substitute 5.47 for t in the original equation and see that h ~ 20m