Problema Solution
A juggler tosses a ball into the air. the ball leaves the juggler's hand 4 feet above the ground and has an initial velocity of 40 feet per sec. the juggler catches the ball when it falls back to a height of 3 feet. use the vertical motion model where h is the height, t is the time in motion, h0 is the initial height, and v is the initial velocity to find how long the ball is in the air.
Answer provided by our tutors
Vertical Motion Model with Initial Velocity is described with:
h(t) = -16t^2 + v*t + h0
h(t) = the height above the ground after t seconds
v = 40 ft/s initial velocity
h0 = 4 ft the initial height of the object (the ball)
t = the time of the motion
h(t) = -16t^2 + 40*t + 4
we need to find t such that h(t) = 3 ft
3 = -16t^2 + 40*t + 4
-16t^2 + 40*t + 4 = 3
by solving we find
t = 2.525 seconds
click here to see the step by step solution of the equation
the ball is approximately 2.525 second in the air.
We can also see the solution by drawing the graph of the function (the graph describes the path of the ball)
y = -16x^2 + 40*x + 4
click here to see the graph