Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s , releases a sandbag at an instant when the balloon is a height h = 40.0 m 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.
A) Compute the position of the sandbag at a time 1.05 s after its release.
B)Compute the velocity of the sandbag at a time 1.05 s after its release.
c) How many seconds after its release will the bag strike the ground?

Answer 1

Data:

Initial velocity upward: Vo = 5.00 m/s ,
Initial position: h = 40.0 m above the ground

Type of motion: free fall.

A) Compute the position of the sandbag at a time 1.05 s after its release.

Equation: y = h + Vo*t - g*(t^2) / 2

y = 40.0 m + 5.00 m/s * 1.05s - (9.8 m/s^2) * (1.05 s)^2 / 2 = 39.8 m

B)Compute the velocity of the sandbag at a time 1.05 s after its release.

Equation: Vf = Vo - g*t

=> Vf = 5.00 m/s - (9.8m/s^2) * (1.05 s) = - 5.29 m/s

Negative sign means that the sandbag is going down.

c) How many seconds after its release will the bag strike the ground?

Equation:

y = yo + Vo*t - g*(t^2) / 2

0 = 40.0 + 5.00t - 4.9 t^2

=> 4.9 t^2 - 5t - 40 = 0

Use the quadratic formula and you get: t = 3.41 s

Answer 2

(a). Position of sandbag at time $1.05\text{ s}$ after its release is $\boxed{39.84\text{ m}}$ above the ground.

(b). Velocity of the sandbag after time $1.05\text{ s}$ is $\boxed{5.3\text{ m/s}}$.

(c). The time taken after release the bag to strike the ground is $\boxed{3.41\text{ s}}$.

Further explanation:

Here, all the actions performed is under free fall. So, we will use the kinematic equations of motion for free falling body.

Given:

The velocity of rising of hot air balloon is $5\text{ m/s}$.

Height of hot air balloon when sandbag released is $40\text{ m}$.

Calculation:

Part (a)

Position of sandbag at time $1.05\text{ s}$ after its release.

When sandbag released the hot air balloon was rising up with the velocity of $5\text{ m/s}$.

So, initial velocity of sandbag will be $5\text{ m/s}$ in upward direction.

So, the time taken by the sand bag to reach at its top position is given by,

$\boxed{v = u - gt}$                                                     …… (1)

Here, $v$ is the final velocity, $u$ is the initial velocity, $g$ is the acceleration due to gravity and negative sign is due upward motion of sandbag, $t$ is the time required to reach at top position.

Substitute values for v and u in equation (1).

$\begin{aligned}0&=5-9.8t\\9.8t&=5\\t&=0.51\text{ s}\\\end{aligned}$

So, the distance travel by sandbag to top position can be calculated as,

$\boxed{{v^2}={u^2}-2g{s_1}}$

Substitute values for $v$ and u in above equation.

$\begin{aligned}{0^2}&={5^2}-2\times9.8\times{s_1}\\19.6{s_1}&=25\\{s_1}&=1.27\text{ m}\\\end{aligned}$

After that sandbag will start falling.

The time remain from the given time is,

$\begin{aligned}{t_1}&=1.05-0.51\\{t_1}&=0.54\text{ s}\\\end{aligned}$

The distance travel by sandbag in $0.54\text{ s}$ in downward direction can be calculated as,

Substitute $0$ for $u$ and $0.54\text{ s}$ for $t$ in above equation.

$\begin{aligned}{s_2}&=0\times0.54+\frac{1}{2}\times9.8{\left({0.54}\right)^2}\\&=1.43\text{ m}\\\end{aligned}$

So, the position of the sandbag after $1.05\text{ s}$ from the ground can be calculated as,

$\begin{aligned}h&=40+{s_1}-{s_2}\\&=40+1.27-1.43\\&=39.84\text{ m}\\\end{aligned}$

Part (b)

Velocity of the sandbag after time $1.05\text{ s}$.

The velocity of the sandbag after time $t$ can be calculated as,

$\boxed{v=u+gt}$

Substitute the values for $u$ and t in above equation.

$\begin{aligned}v&=0+9.8\times0.54\\&=5.3\text{ m/s}\\\end{aligned}$

Thus, the velocity of the sandbag after time $1.05\text{ s}$ is $\boxed{5.3\text{ m/s}}$.

 

Part (c)

The time taken after release the bag to strike the ground.

The total distance the top most position of the bag and the ground is,

$\begin{aligned}S&=40+{s_1}\\&=40+1.27\\&=41.27\text{ m}\\\end{aligned}$

Now, time taken by the bag to strike the ground from its top most position,

$\boxed{S=ut+\dfrac{1}{2}g{t_2}^2}$

 

Substitute $41.27{\text{ m}}$ for $S$ and $0$ for $u$ in above equation.

$\begin{aligned}41.27&=0\timest+\dfrac{1}{2}\times9.8{t_2}^2\\41.27&=4.9{t_2}^2\\{t_2}^2&=\dfrac{{41.27}}{{4.9}}\\&=2.9{\text{ s}}\\\end{aligned}$

Now, the total time taken by bag to strike the ground from the instant of release is,

$\begin{aligned}T&={t_2}+t\\&=2.9+0.51\\&=3.41\text{ s}\\\end{aligned}$

Thus, the time taken after release the bag to strike the ground is $\boxed{3.41\text{ s}}$.

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Answer detail:

Grade: Senior School

Subject: Physics

Chapter: Kinematics

Keywords:

Hot air balloon, constant velocity, height of, position of sandbag, velocity of sandbag, total time, rising up, 5m/s, 40m, 1.05 s, displacement, balloonist, strike the ground.