A hotair balloonist, rising vertically with a constant speed of 5.00 m/s, releas

Question

A hotair balloonist, rising vertically with a constant speed of 5.00 m/s, releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no appreciable air drag.

A.)Compute the position (height above the ground) of the sandbag at 0.272 after its release.
B.)Compute the position of the sandbag 1.09 after its release.
C.)Compute the component of the velocity of the sandbag at 0.272 after its release.
D.)Compute the component of the velocity of the sandbag at 1.09 after its release.

Explanation / Answer

Okay so you start with: a = -g Integrate this: v = -gt + v0 Integrate this: x = -g/2(t^2) + v0*t + x0 Where x = height from ground, t = time, v0 = starting velocity, x0 = starting distance from ground, v = velocity, and a = accelleration. From this you can get all of your answers: A) x(t = .272) = (-4.9)*(.272^2) + (5.00*0.272) + 40m = 40.997m B) x(t = 1.09) = (-4.9)*(1.09^2) + (5.00*1.09) + 40m = 39.628m C) v(t = .272) = (-9.8*.272) + 5.00 = 2.334 m/s D) v(t = 1.09) = (-9.8*1.09) + 5.00 = -5.682 m/s

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