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an object is shot straight upward from sea level with an initial velocity of 250 ft/sec.
a)assuming that gravity is the only force acting on the object, give an upper estimate for it's velocity after 5 seconds has elapsed. use g=32 ft/sec^2 for gravitational acceleration.
b) using delta t=1 sec, find a lower estimate for the height attained after 5 seconds. ...?

Respuesta :

syed514
 There are exact formulas for what you have asked for. 
a) V = V0 + g*t 
So V = 400 + -32 * 5 = 400 - 160 = 340 ft/sec 
b) s = s0 + v0*t + (1/2)*g*t^2 
s = 0 + 400*5 - 16*25 = 2000 - 400 = 1600 ft 

You do not state in your problem how you are to estimate these.
obasola1

Answer:

.a 90 ft/sec     b. s=400ft

Explanation:

when an object is shot straight upward, the force of gravity which is the earth magnetic pull  on an object to its center, acts on it

The basic parameters to be considered are

a=-g, -32ft/sec

s=distance/height attained by the object during its motion against gravity

t=time ,secs

u=initial velocity

V=final velocity ft/sec

from newtons equation of motion,

a) V = u + g*t  

So V = 250+ -32*5= 90 ft/sec  

it's velocity after 5 seconds has elapsed is  90 ft/sec  

b) s = [tex]s=ut+\frac{1}{2} *at^{2}[/tex]

[tex]s=250*5+\frac{1}{2}-32 *5^{2}[/tex]

s=400ft

the height attained after 5 seconds. ..s=400ft

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