A hotair balloonist, increasing vertically v a continuous speed of 5.00m/s , releases a sandbag at the prompt the balloon is 40.0m over the ground. After it is released, the sandbag encounters no appreciable air drag.

You are watching: How many seconds after its release will the bag strike the ground?

A) Compute the position (height above the ground) the the sandbag at 0.251s ~ its release.

B) Compute the place of the sandbag 1.07s ~ its release.

C) Compute the x ingredient of the velocity that the sandbag in ~ 0.251s ~ its release.

D) Compute the x ingredient of the velocity that the sandbag at 1.07s after its release.

E) How many seconds after its release will the bag to win the ground?

F) How quick is it moving as it strikes the ground

G) What is the biggest height above the ground the the sandbag reaches?  edit Treebeard
31978
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1)

here, u= -5, g=9.81, t=.251

s= ut + (1/2) gt^2

= -5*.251 + .5*9.81*.251^2

= -0.9459

height of sandbag above ground in ~ .262s is 40+.9459=40.9459 m

2)

s= ut + (1/2) gt^2

= 5*1.07 + .5*9.81*1.07^2

= 0.2657

height of sandbag over ground at .262s is 40-0.2653=39.7343 m

3)

here, u =-5, t=.251, g=9.81, v=?

v=u+gt

= -5+9.81*.251

= -2.537 m/s

velocity that sandbag at .251 s after relax is 2.537 m/s upwards.

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4)

t=1.07 s

v=u+gt

= -5 + 9.81*1.07

= 5.496

velocity of sandbag at 1.07 s after relax is 5.496 m/s downwards.

5)

u=-5, h=40, g=9.81, t=?

h= ut + .5 gt^2

40= -5t + .5*9.81 t^2

4.905t^2-5t-40=0

this quadratic equation offers two services t=3.41 and also -2.39

accepting just positive value, t=3.41 s

6)

u=-5, t=3.41s, g=9.81, v=?

v=u+gt

=-5 + 9.81*3.41

=28.4521

at the minute it strikes ground, v=28.4521 m/s

7)

u=5, v=0, g=-9.81, h=?(in this situation upward being positive)

v^2 = u^2 + 2gh

0=5^2 - 2*9.81*h

h=1.274m

greatest height above ground 40+h=41.274m  edit Treebeard
31978
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