kinematics: How long does the can take to fall to the ground

Question: A hot air balloon is rising upward with a constant velocity of 4.0 m/s. As the balloon reaches a height of 4.0 m above the ground, the balloonist accidently drops a can of pop over the edge of the basket. How long does it take the pop can to reach the ground?

So,

v= 4.0 m/s

d= 4.0 m

a= 9.8 m/s (squared)

As was explained in your other kinematics thread, you need to plug the given information into the standard "projectile motion" formula, s(t) = -4.9t² + v₀t + h₀.

You've been given v₀ and h₀, and have been asked to find t when s(t) = 0. So plug in zero for s(t) and solve.

Eliz.

Re: kinematics: How long does the can take to fall to the gr

Maryanne said:

Question: A hot air balloon is rising upward with a constant velocity of 4.0 m/s. As the balloon reaches a height of 4.0 m above the ground, the balloonist accidently drops a can of pop over the edge of the basket. How long does it take the pop can to reach the ground?

So,

v= 4.0 m/s

d= 4.0 m

a= 9.8 m/s (squared)

Click to expand...

When released, the can has an upward velovity of 4m/s
From Vf = Vo - gt, or 0 = 4 - 9.8t, or t = 0.4081sec.

During that time period, the can continued to rise until it reached zero verticalvelocity or h = Vot - 9.8t^2/2 or h = 4(0.4081) - 4.9(0.4081)^2 or
h = 0.816m.

Npw the can falls from a height of 4.816m with Vo = 0.

From h = Vot + 4.9t^2 we have 4.816 = 0 + 4.9t^2 making t^2 = 0.9829 or t = 0.9914sec. to reach the groung.