Step 1

1Concepts and Principles

1- Particle under constant acceleration: If a particle moves in a straight line with a constant acceleration $a}_{x$, its motion is described by kinematics equations, from which we will use the following equation:

(1)${v}_{xf}={v}_{xi}+{a}_{y}t$

(2)$x}_{f}={x}_{i}+{v}_{\xi}t+\frac{1}{2}{a}_{x}{t}^{2$

2- The sulution to a quadration in the from: $a{x}^{2}+bx+x=0$ is given by the quadrat formula:

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

Step 2

2 Given Data

Let the +y-direction be upwards and the initial position of the sand bag be at ${y}_{i}=0$.

$v}_{yi}\text{(initial velocity of the bag)}=5\frac{m}{s$

${y}_{f}\text{(position of the bag when it reaches the ground)}=-40m$

$a}_{y}\text{(acceleration due to gravity)}=-9.8\frac{m}{{s}^{2}$

Step 3

3 Required Data

In part (a), we are asked to find the position and the velocity of the sandbag at t=0.25 s and t=1 s.

In part(b), we are asked to find the time taken by the bag to reach the ground.

In part(c), we are asked to find the speed with which the bag strikes the ground.

In part(d), we are asked to find the maximum height the bag reaches above the ground.

In part(e), we are asked to sketch ${a}_{y}-t,{v}_{y}-t,\text{and}\text{}y$ task graphs for the motion of the sand bag.

Step 4

4 Solution

(a)

Model the sand bag as a particle under constant acceleration and apply Equations (1) and (2) to find the final velocity and final position of the sand bag as a function of time:

${v}_{yf}={v}_{yi}+{a}_{y}t$

$y}_{f}={y}_{i}+{v}_{yi}t+\frac{1}{2}{a}_{y}{t}^{2$

For t=0.25 s:

$v}_{yf}=5\frac{m}{s}+(-9.8\frac{m}{{s}^{2}})\left(0.25s\right)=2.55\frac{m}{s$

${y}_{f}=0+\left(5\frac{m}{s}\right)\left(0.25s\right)+\frac{1}{2}(-9.8\frac{m}{{s}^{2}}){\left(0.25s\right)}^{2}=0.944m$

Therefore, at t=0.25 s, the sand bag is moving with an upward velocity of magnitude 2.55 m/s and is 40 m+0.944 m=40.9 m above the ground.

For t=0.1 s:

$v}_{yf}=5\frac{m}{s}+(-9.8\frac{m}{{s}^{2}})\left(1s\right)=-4.8\frac{m}{s$

${y}_{f}=0+\left(5\frac{m}{s}\right)\left(1s\right)+\frac{1}{2}(-9.8\frac{m}{{s}^{2}}){\left(1s\right)}^{2}=0.1m$

Therefore, at t=1 s, the sand bag is moving with an downward velocity of magnitude 4.8 m/s and is 40 m+0.1 m=40.1 m above the ground.