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Answer :
Answer:
(a) -32ft/s
(b) 12.85s
(c) -411.2ft/s
Explanation:
The equation of the position, s(t), of the object at any time t is given as;
s(t) = -16t² + 181 ---------------------(i)
(a) To calculate the velocity of the object, first take the derivative of equation (i) with respect to t - since the change in position with respect to time t gives the velocity. This will give the velocity function, v(t), of the object at any time t. i.e
v(t) = [tex]\frac{ds(t)}{dt}[/tex]
v(t) = -32t ---------------------(ii)
The velocity of the object 1 second after being dropped can be found by substituting t=1 into equation (ii) as follows;
v(1) = -32(1)
v(1) = -32
Therefore, the velocity after 1 second is -32 ft/s.
Note: The above result is in ft/s as the position is measure in feet above the ground.
(b) To calculate the time it will hit the ground, first observe that at ground level (when it hits the ground), the position of the object will be 0 relative to the ground. Therefore, substitute s(t) = 0 into equation (i) to get the time it will take to hit the ground as follows;
s(t) = -16t² + 181
0 = -16t² + 181
16t² = 181
t² = 181 / 16
t² = 165
Solve for t;
t = [tex]\sqrt{165}[/tex]
t = 12.85s
Therefore, the time it will take to hit the ground is 12.85s
(c) To calculate its velocity upon impact (upon hitting the ground), substitute the value of the time taken to hit the ground (12.85), into the velocity's equation - equation (ii) as follows;
v(t) = -32t
v(12.85) = -32(12.85)
= -411.2m/s
Therefore, the velocity of the object upon impact is -411.2 ft/s
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Rewritten by : Barada
Answer:
a. - 32
b. 3.36
c. -107.6
Explanation:
[tex]s(t) = -16t^2 + 181[/tex]
a. The velocity is obtained as the time-derivative of the displacement.
[tex]v(t) = -32t[/tex]
At [tex]t=1[/tex],
[tex]v(1) = -32\times1=-32[/tex]
b. The object hits the ground when the height is 0.
[tex]0 = -16t^2 + 181[/tex]
[tex]16t^2 = 181[/tex]
[tex]t^2 = \sqrt{\dfrac{181}{16}} = \dfrac{\sqrt{181}}{4} = 3.36[/tex]
c. It's velocity upon impact is the velocity when [tex]t[/tex] is the time it hits the ground.
[tex]v(t) = -32t[/tex]
[tex]v(\dfrac{\sqrt{181}}{4}) = -32\times\dfrac{\sqrt{181}}{4} = -8\sqrt{181} = -107.6 [/tex]
