Answer
a) The position and velocity of the sandbag from the ground at $t=0.250s$ is $40.9m$ and $2.55\frac{m}{s}$.
The position and velocity of the sandbag from the ground at $t=1.00s$ is $40.1m$ and $-4.80\frac{m}{s}$.
b) The sandbag strikes the ground $3.41s$ after its release.
c) The magnitude of velocity with which the sandbag strikes the ground is $28.4\frac{m}{s}$
d) The maximum height above the ground which the sandbag reaches is $41.3m$.
e) See the graph below.
Work Step by Step
We take the ground to be the origin of our system with (+)ve y-direction upwards and let us start measuring $(t=0)$ when we drop the sandbag at a height of $40.0m$ above the ground .
Since the motion is uniform, we can use the kinematics equations.
a) Let us calculate the position and velocity of the sandbag at $t=0.250s$ and $t=1.00s$.
i) At $t=0.250s$
$y-y_{0}=v_{0y}t+\frac{1}{2}a_{y}t^2$
$y=40.0+5.00\times0.250-\frac{1}{2}\times9.80\times0.250^2=40.9m$
and $v_{y}=v_{0y}+a_{y}t=5.00-9.80\times0.250=2.55\frac{m}{s}$
The position and velocity of the sandbag from the ground at $t=0.250s$ is $40.9m$ and $2.55\frac{m}{s}$.
ii) At $t=1.00s$
$y-y_{0}=v_{0y}t+\frac{1}{2}a_{y}t^2$
$y=40.0+5.00\times1.00-\frac{1}{2}\times9.80\times1.00^2=40.1m$
and $v_{y}=v_{0y}+a_{y}t=5.00-9.80\times1.00=-4.80\frac{m}{s}$
The position and velocity of the sandbag from the ground at $t=1.00s$ is $40.1m$ and $-4.80\frac{m}{s}$
b) Let us now calculate the time taken by the sandbag to strike the ground after its release.
$y-y_{0}=v_{0y}t+\frac{1}{2}a_{y}t^2$
$=>\frac{1}{2}\times9.80t^2-5.00t-40.0=0$
Using quadratic formula, we get
$t=\frac{-(-5.00)+-\sqrt ((-5.00)^2-4\times\frac{9.80}{2}\times(-40.0))}{2\times\frac{9.80}{2}}=3.413s\approx3.41s$
[We take the (+)ve time]
The sandbag strikes the ground $3.41s$ after its release.
c) Let us calculate the velocity with which the sandbag strikes the ground.
$v_{y}=v_{0y}+a_{y}t=5.00-9.80\times3.413=-28.4\frac{m}{s}$
The magnitude of velocity with which the sandbag strikes the ground is $28.4\frac{m}{s}$
d) Let us calculate the maximum heigth above the ground which the sandbag reaches.
$2a_{y}(y-y_{0})=v_{y}^2-v_{0y}^2$ , $v_{y}=0$ at peak height
$y=y_{0}-\frac{v_{0y}^2}{2a_{y}}=40.0-\frac{5.00^2}{-2\times9.80}=41.3m$
The maximum height above the ground which the sandbag reaches is $41.3m$ .
e)See the graph of $a_{y}-t, v_{y}-t$ and $y-t$.
