Respuesta :
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
Using movement concepts, it is found that:
- The object hits the ground after 5.59 seconds.
- The object hits the ground at a velocity of -178.88 feet per second.
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The height of the object after t seconds, dropped from a height of 500 feet, is given by:
[tex]h(t) = -16t^2 + 500[/tex]
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It hits the ground when [tex]h(t) = 0[/tex], thus:
[tex]h(t) = 0[/tex]
[tex]-16t^2 + 500 = 0[/tex]
[tex]16t^2 = 500[/tex]
[tex]t^2 = \frac{500}{16}[/tex]
[tex]t = \sqrt{\frac{500}{16}}[/tex]
[tex]t = 5.59[/tex]
The object hits the ground after 5.59 seconds.
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The velocity is the derivative of the position, thus:
[tex]v(t) = h^{\prime}(t) = -32t[/tex]
The velocity when it impacts the ground is v(5.59), thus:
[tex]v(5.59) = -32(5.59) = -178.88[/tex]
The object hits the ground at a velocity of -178.88 feet per second.
A similar problem is given at https://brainly.com/question/14516604
