Question

The altitude of an object, d, can be modeled using the equation below. d=-16t^2+vh+h.where v is the initial velocity and h is the initial height.A catapult launches a stone with an initial velocity of 48 feet per second and an initial height of 0 feet. How long will the stone be in flight?

Answer 1

The stone will be in flight for 3 seconds

Explanation:

Given the equation;

d = -16t^2 + vt + h

Where;

v = the initial velocity

h = the initial height

Given;

v = 48 ft/s

h = 0

Substituting into function d;

d = -16t^2 + 48t + 0

d = -16t^2 + 48t

At the point of launch and the point of landing d = 0.

We need to calculate the values of t for d to be equal to zero.

d = -16t^2 + 48t = 0

-16t^2 + 48t = 0

Factorising, we have;

16(-t+3)t = 0

So,

t = 0 or -t+3 = 0

t= 0 or t =3

Change in t is;

∆t = t2 - t1 = 3 - 0

∆t = 3 seconds.

The stone will be in flight for 3 seconds.