Solving a word problem

Question

A ball Is thrown from a height of 205' with an initial downwards velocity of 11ft/s the ball's height h in feet after t seconds is given by the following h=205 - 11t -16t^2 how long after the ball is thrown does hit the ground

Answer 1

about 3 1/4 seconds

set h=0, solve for t

-16t² -11t +205 = 0

If you can't factor the left side, then use the quadratic formula

t=(11 + or - (121-4(-16)(205))^1/2)/2(-16)

Ignore the positive square root as it leads to negative time

11 + the negative square root of 13241 all divided by -32 = about 3.25 seconds

Just the initial downward speed of 11ft/s would get the ball to ground level in 205/11 = about 19 seconds

but the time is far less due to gravity increasing the speed 32 ft/s/s. On average it will travel about

5 times as fast as the initial velocity, with t = about 1/5 of 19 seconds or less than 4 seconds

Or if you prefer graphs, graph a few points of the parabola. Plug simple values into the original h equation

It's maximum point is found taking the 1st derivative, -32t-11=0 t=about -1/3 seconds

at time t=0 h=205

t=1 h=178

t=2 h=119

t=3 h=28

t=4 h=-84

h=0 somewhere between t=3 and t=4

0 is much closer to 28 than to -84, so t= 3+ seconds, but closer to 3 than to 4

84/28=3 the time for h=0 is 3 times closer to t=3 than to t=4. It's one 1/4 of

the total distance from 28 to -84, so 1/4 of the way from 3 to 4 is 3.25 seconds

carefully plot the points, graph it with a smooth parabolic curve through the points and

look where it intersects the x-axis. It is close to the point (3.25, 0)