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 th e 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?
1 answer:
Answer:
The stone will be in flight for 3 seconds
Step-by-step 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.
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