Problema Solution
A person Spikes the volleyball over a net when the ball is 9 feet above the ground. The volleyball has an initial vertical velocity of -40 ft./s. The volleyball is allowed to fall to the ground. How long is it on the air after it is spiked?
Answer provided by our tutors
The height h (in(feet) of an object after it is launched is given by the function:
h(t) = -16t^2 + v0t + h0,
where v0 is the initial velocity and in our case is v0 = -40 ft/s
ho is the initial height of the object and in hour case h0 = 9 ft
t is the time in seconds after the object is launched
h(t) = -16t^2 - 40t + 9,
we need to find t, such that t>0, and h(t) = 0 that is
-16t^2 - 40t + 9 = 0
by solving we find:
t = 0.21 s
click here to see the step by step solution of the equation:
the ball is in the air for 0.21 seconds.