A hot-air balloonist, rising vertically with a constant velocity of magnitude v

Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00m/s . releases a sandbag at an instant when the balloon is a height h = 40.0m above the ground (Figure 1). After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive. Compute the position of the sandbag at a time 0.245s after its release. Compute the velocity of the sandbag at a time 0.245s after its release. Compute the position of the sandbag at a time 1.50s after its release.

Explanation / Answer

here, u= -5, g=9.81, t=.245 s= ut + (1/2) gt^2 = -5*.262 + .5*9.81*.245^2 = -.96 height of sandbag above ground at .245s is 40+.96=40m 2. here, u =-5, t=.245, g=9.81, v=? v=u+gt = -5+9.81*.245 = -2.3 m/s velocity of sandbag at .245 s after release is 2.3 m/s upwards. 3. t=1.5 s v=u+gt = -5 + 9.81*1.5 = 9.715 velocity of sandbag at 1.5 s after release is 9.7165 m/s downwards. 4. 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 gives two solutions t=3.41 and -2.39 accepting only positive value, t=3.41 s 5. u=-5, t=3.41s, g=9.81, v=? v=u+gt =-5 + 9.81*3.41 =28.4521 at the moment it strikes ground, v=28.4521 m/s 6. u=5, v=0, g=-9.81, h=?(in this case 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

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