Answer :
Answer:
ΔS = 1/2gt²
Step-by-step explanation:
Given
Initial height of the ball = 3 feet
Maximum height of the ball = 403 feet
To get the function that best suits the situation after t seconds, we will use the equation of motion;
ΔS = ut + 1/2gt²
ΔS is the change in height
u is the initial velocity of the body = 0m/s
g is the acceleration due to gravity = 9.81m/s²
t is the time taken for the ball to reach its maximum height.
ΔS = 403-3
ΔS = 400ft
Substituting the given values into the formula;
ΔS = ut + 1/2gt²
400 = 0 + 1/2(9.81)t²
400 = 4.905t²
t² = 400/4.905
t² = 81.55
t = √81.55
t = 9.03secs
The situation that best model the situation is ΔS = ut + 1/2gt²
ΔS = 0 + 1/2gt²
ΔS = 1/2gt²