(a) Yes
We can calculate the total vertical distance covered by the rock by using the equation
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final velocity of the rock (the maximum height is reached when v=0)
u = 8.60 m/s is the initial velocity
a = g = -9.8 m/s^2 is the acceleration due to gravity (negative since it points opposite to the velocity)
d is the vertical distance covered
Solving for d,
[tex]d=\frac{v^2 -u^2}{2g}=\frac{0-(8.60 m/s)^2}{2(-9.8 m/s^2)}=3.77 m[/tex]
The rock starts its motion from a heigth of
[tex]y_0 = 1.59 m[/tex]
So the maximum height it will reach is
[tex]h=y_0 + d=1.59 m +3.77 m =5.36 m[/tex]
and since the wall is 3.62 m high, this means that the rock will reach the top of the wall.
(b) 5.85 m/s
We can find the velocity of the ball when it reaches the top of the wall by using again the formula
[tex]v^2 - u^2 = 2gd[/tex]
where this time we have
v = ? is the velocity of the rock at the top of the wall
u = 8.60 m/s is the initial velocity
g = -9.8 m/s^2
d = 3.62 m - 1.59 m = 2.03 m is the vertical distance covered by the rock to reach the top of the wall
Solving for v,
[tex]v=\sqrt{u^2 + 2gd}=\sqrt{(8.60 m/s)^2+2(-9.8 m/s^2)(2.03 m)}=5.85 m/s[/tex]
(c) 2.07 m/s
Again, we can use the same formula:
[tex]v^2 - u^2 = 2gd[/tex]
But this time we have:
v = ? is the final velocity of the rock at the bottom of the wall
u = 8.60 m/s is the initial velocity of the rock
g = +9.8 m/s^2 , this time positive since the acceleration (downward) has same direction as the velocity (downward)
d = 2.03 m is the distance between the top of the wall and the final point of the motion (1.59 m above the ground)
Solving for v,
[tex]v=\sqrt{u^2 + 2gd}=\sqrt{(8.60 m/s)^2+2(+9.8 m/s^2)(2.03 m)}=10.67 m/s[/tex]
So, the change in speed is
[tex]\Delta v = v -u =10.67 m/s - 8.60 m/s = 2.07 m/s[/tex]
(d) No
In fact, the change in speed of the rock moving upward is (using the result found in part b):
[tex]\Delta v = v-u = 5.85 m/s - 8.60 m/s = -2.75 m/s[/tex]
which we see is different from the change in speed found at part c (rock moving downwards).
This is due to the fact that the rock has an initial velocity different from zero: for this reason, the two situations are not symmetrical:
- when the rock is thrown upward, the gravity acts against the direction of motion, so it makes the rock slowing down
- when the rock is thrown downward, the gravity acts in the same direction as the motion, so it makes the rock speeding up
