A fireworks shell is shot upward with an initial velocity of 28m/s from a height of 2.5 m.
s(t) = -4.9t^2 + 28t + 2.5
After how many seconds will the shell have the same speed, but be falling downward?
v(t) = -9.8t + 28
we want this to be -28 (negative for downwards)
-9.8t + 28 = -28
-9.8t = -56
t = -56/-9.8 = 5.7 seconds
To find out after how many seconds the shell will have the same speed but be falling downward, we need to determine when the velocity of the shell is equal to zero.
The velocity, v(t), can be obtained by taking the derivative of the position function, s(t), with respect to time:
v(t) = ds(t)/dt
Given the position function:
s(t) = -4.9t^2 + 28t + 2.5
Taking the derivative:
v(t) = d/dt (-4.9t^2 + 28t + 2.5)
v(t) = -9.8t + 28
Now we set the velocity equal to zero and solve for t:
-9.8t + 28 = 0
-9.8t = -28
t = -28 / -9.8
t ≈ 2.86 seconds
Therefore, after approximately 2.86 seconds, the shell will have the same speed, but be falling downward.