SOLUTION: the height h, of a ball above ground t seconds after it is thrown vertically upwards can be approximated by the formula h=vt-5t squared, wher v is the intial speed with which the b

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: the height h, of a ball above ground t seconds after it is thrown vertically upwards can be approximated by the formula h=vt-5t squared, wher v is the intial speed with which the b      Log On


   

Question 35777: the height h, of a ball above ground t seconds after it is thrown vertically upwards can be approximated by the formula h=vt-5t squared, wher v is the intial speed with which the ball is released in meters per second.
at what two times will a ball be 105m above ground if it is thrown vertically upwards with an initial speed of 50m/s
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
h+=+vt-5t%5E2
Substitute v = 50 and h = 105 and solve for t.
105+=+50t+-+5t%5E2 Subsubtract 105 from both sides and simplify.
5t%5E2+-+50t+%2B+105+=+0 Solve this quadratic equation by factoring.
5%28t%5E2+-+10t+%2B+21%29+=+0 Apply the zero products principle.
t%5E2+-+10t+%2B+21+=+0 Factor.
%28t+-+3%29%28t+-+7%29+=+0 Apply the zero products principle again.
t+-+3+=+0 and t+-+7+=+0
If t+-+3+=+0 then t+=+3
If t+-+7+=+0 then t+=+7
The two times are:
t = 3 seconds.
t = 7 seconds.
You check these solutions by substituting into the original equationh+=+50t-5t%5E2 and solving for h.