Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only force acting on it thereafter is gravity. Take g = 32 ft/sec2 . (a) Find the highest altitude attained by the object. (b) Determine how long it takes for the object to fall to the ground.

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec²  [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

vapour or gas assuming atmospheriv pressure

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures