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:
tan−1(StartFraction 6.9 Over 9.8 EndFraction)
Step-by-step explanation:
tan−1(StartFraction 6.9 Over 9.8 EndFraction)
tan = opp/adj = 9.8/6.9
tan -1 = 1 / tan = 1 / (9.8 /6.9) = 6.9 /9.8
For this problem, we plug in the <em>numbers for t</em> and the <em>appropriate letters for e</em> to the equation <em>e = 300 - 10t</em>.
a) a = 300 - 10(-2)
simplify: a = 300 + 20
simplify: a = 320 ft.
But, this is not a viable point because when t≤0, she doesn't move anywhere, thus, should consistently be 300 ft.
b) b = 300 - 10(3.5)
simplify: b = 300 - 35
simplify: b = 265 ft.
c) c = 300 - 10(30)
simplify: c = 300 - 300
simplify: c = 0 ft.