The zeros are the values of t for which f(t) = 0.
i.e. <span>-16t^2 + 96t = 0
16</span>t^2 - 96t = 16t(t - 6)
16t = 0 or t - 6 = 0
t = 0 or t = 6
Therefore, the zeros are 0, 6
The time taken for the ball to hit the ground is the value of t when f(t) = 0.
i.e. t = 6.
Answer:
160m/s
Step-by-step explanation:
The object can hit the ground when t = a; meaning that s(a) = s(t) = 0
So, 0 = -16a² + 400
16a² = 400
a² = 25
a = √25
a = 5 (positive 5 only because that's the only physical solution)
The instantaneous velocity is
v(a) = lim(t->a) [s(t) - s(a)]/[t-a)
Where s(t) = -16t² + 400
and s(a) = -16a² + 400
v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)
v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)
v(a) = lim (t->a) -16(t² - a²)(t-a)
v(a) = -16lim t->a (t²-a²)(t-a)
v(a) = -16lim t->a (t-a)(t+a)/(t-a)
v(a) = -16lim t->a (t+a)
But a = t
So, we have
v(a) = -16lim t->a 2a
v(a) = -32lim t->a (a)
v(a) = -32 * 5
v(a) = -160
Velocity = 160m/s
Answer:
2.5
Step-by-step explanation:
