Answer
(a) At $t=2.04s$, the boulder is moving at $20.0\frac{m}{s}$ upwards.
(b) At $t=6.12s$, the boulder is moving at $20.0\frac{m}{s}$ downwards.
(c) At $t=8.16s$, the displacement of the boulder from its initial position is 0.
(d) At $t=4.08s$, the velocity of the boulder is 0.
(e) The magnitude and direction of acceleration is $9.80\frac{m}{s^2}$ downwards throughout the motion.
(f) See the sketch.
Work Step by Step
Let the initial position of the boulder be the origin of our system with (+)ve y-direction upwards.
Since air resistance is ignored, we can consider the motion of the boulder to be free-fall motion. So we can apply the kinematics equation for constant acceleration.
The initial speed of the boulder is given to be $v_{i}=40.0\frac{m}{s}$.
(a) We want to find the time at which the boulder is moving at $v_{f}=20.0\frac{m}{s}$ upwards.
$v_{f}=v_{i}+at$ , where a=acceleration due to gravity
$t=\frac{v_{f}-v_{i}}{a}=\frac{20.0-40.0}{-9.80}=2.04s$
At $t=2.04s$, the boulder is moving at $20.0\frac{m}{s}$ upwards.
(b) Let us calculate the time t the boulder is moving at $v_{f}=-20.0\frac{m}{s}$ downwards.
$t=\frac{v_{f}-v_{i}}{a}=\frac{-20.0-40.0}{-9.80}=6.12s$
At $t=6.12s$, the boulder is moving at $20.0\frac{m}{s}$ downwards.
(c) We now want to calculate the time t when the boulder returns to the initial position (assuming the motion to be purely vertical).
When the boulder returns to its initial position the boulder is moving at $40.0\frac{m}{s}$ downwards.
$t=\frac{v_{f}-v_{i}}{a}=\frac{-40.0-40.0}{-9.80}=8.16s$
At $t=8.16s$, the displacement of the boulder from its initial position is 0.
(d) The velocity of the boulder is 0 at peak height of the motion.
$t=\frac{v_{f}-v_{i}}{a}=\frac{0-40.0}{-9.80}=4.08s$
At $t=4.08s$, the velocity of the boulder is 0.
(e) Since the motion of the boulder is free-fall motion, the acceleration is due to gravity .
The magnitude and direction of acceleration due to gravity is constant throughout the motion.
So, the magnitude and direction of the acceleration when the boulder is (i) moving upward (ii) moving downward (iii) at the highest point is $9.80\frac{m}{s^2}$ downwards.
(f) See the sketch of $a_{y}-t$, $v_{y}-t$ and $y-t$ for the motion of the boulder.
