If an object is dropped from a 181-foot-high building, its position (in feet above the ground) is given by s(t)equalsminus16tsquaredplus181, where t is the time in seconds since it was dropped. a. What is its velocity 1 second after being dropped? b. When will it hit the ground? c. What is its velocity upon impact?

a. - 32

b. 3.36

c. -107.6

Explanation:

a. The velocity is obtained as the time-derivative of the displacement.

At ,

b. The object hits the ground when the height is 0.

c. It's velocity upon impact is the velocity when is the time it hits the ground.

(a) -32ft/s

(b) 12.85s

(c) -411.2ft/s

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) =

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 =

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

answera compound;

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j