Question

A hot air balloonist, rising vertically with a constant velocity of magnitude 5 m/s, releases a sandbag at an instant when the balloon is 40 m above the ground. after it release, the sandbag is in free fall.
a) Complete the position and velocity of the sandbag at 0.250s and 1.00s after its release .
b) How many seconds after its release will the bag strike the ground ?
c) What is the magnitude of velocity does it sytrike the ground? (d) what is the greatest height above the ground that the sandbags reaches (ay-t), vy-t and y-t graph for the motion

Answer 1

$a=g=9.8 m/s \\ Y_0=40m\\ V_0=5m/s$

a) $V=V_0-at$

$Y = Y_0+V_0t-at^2/2$

velocity at $t=0.250s:$

$V=5-9,8*0.250=2.55 \\ V(t=0.250s)=2.55 m/s$

position at $t=0.250s:$

$Y = 40+5*0.250-\frac{9.8}{2}*0.25^2\\ Y(t=0.250s)=40.94 m$

velocity at $t=1.00s:$

$V=5-9,8*1.00=4.8 \\ V(t=1.00s)=-4.8 m/s$

position at $t=1.00s:$

Y(t=1.00s)=40.1 m

$Y = 40+5*1.00\frac{9.8}{2}*1.00^2\\ Y(t=1.00s)=40.1 m$

b)  bag strike the ground at Y=0

$Y =40+5t-4.9t^2\\ 40+5t-4.9t^2=0\\ t_1=3.41 \\$

$t_2=-2.3921$ - the time can't be below zero

$t=3.41s$

c)

$V=5-9,8*3.41=28.44 \\ V(t=3.41s)=28.44 m/s$

d) $Y_{max}$ if $V=0$

$V=5-9.8t\\ 5-9.8t=0\\ t=0.51\\ Y=40+5*0,51-4.9*0.51^2=41.27\\ Y_{max}=41.27 m$