SOLUTION: A ball is thrown vertically upward with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=48t−16t2
.
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: A ball is thrown vertically upward with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=48t−16t2
.
Log On
Question 1206150: A ball is thrown vertically upward with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=48t−16t2
.
(a) At what time t will the ball strike the ground?
(b) For what time t is the ball more than 20 feet above the ground? Found 2 solutions by mananth, greenestamps:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! A ball is thrown vertically upward with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=48t−16t2
.
(a) At what time t will the ball strike the ground?
at t=0 the ball is on ground
S(t)=0 = 48t-16t^2
Solve for t
16t^2=48t
t=3s
(b) For what time t is the ball more than 20 feet above the ground?
Between two times it is above 20 ft
48t-16t^2 >20
Solve the inequality to get the time (s)
.
(a) The ball hits the ground when s, the height of the ball above the ground, is zero:
and
ANSWER: at 3 seconds
(b) We want the height of the ball to be greater than 20 feet:
It's awkward to solve the quadratic inequality. Instead, solve an equation to find the times when the height of the ball is exactly 20 feet. Then use common sense to get the answer.
and
The ball is 20 feet off the ground at t=0.5 seconds and t=2.5 seconds, so it is more than 20 feet above the ground between those two times.
.
