c) A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff the rock has the speed of 29 m/s. Assuming that the air resistance can be ignored, find
i. the initial speed of the rock
(5 marks)
ii. the greatest height of the rock as measured from the base of the cliff.
Final KE=initial PE+initial KE
solve fodr initial KE, then vi
greatest height?
mgh=finalKE
25.18m/s
To find the initial speed of the rock (i), we can use the equation for the conservation of energy. The total mechanical energy at the top of the cliff is equal to the total mechanical energy at the bottom.
At the top of the cliff:
Potential Energy (PE) = mgh (where m is the mass of the rock, g is the acceleration due to gravity, and h is the height of the cliff)
Kinetic Energy (KE) = 0 (the rock is initially at rest)
At the bottom of the cliff:
Potential Energy (PE) = 0 (the rock is at ground level)
Kinetic Energy (KE) = 1/2 mv^2 (where v is the final speed of the rock)
Using the conservation of energy equation:
PE_top + KE_top = PE_bottom + KE_bottom
mgh + 0 = 0 + 1/2 mv^2
Simplifying the equation:
gh = 1/2 v^2
Now, plug in the given values:
m = 0.26 kg
g = 9.8 m/s^2
h = 32 m
v = 29 m/s
Solving for the initial speed (i):
i. The initial speed of the rock is the value we are looking for.
Plug in the values into the equation:
(9.8 m/s^2)(32 m) = 1/2 (29 m/s)^2
313.6 = 1/2 (841)
313.6 = 420.5
Now, multiply both sides of the equation by 2:
2 * 313.6 = 2 * 420.5
627.2 = 841
Therefore, the equation is not balanced.
Correcting,
Using the conservation of energy equation:
PE_top + KE_top = PE_bottom + KE_bottom
mgh + 0 = 0 + 1/2 mv^2
Simplifying the equation:
gh = 1/2 v^2
Now, plug in the given values:
m = 0.26 kg
g = 9.8 m/s^2
h = 32 m
v = 29 m/s
Solving for the initial speed (i):
i. The initial speed of the rock is the value we are looking for.
Plug in the values into the equation:
(9.8 m/s^2)(32 m) = 1/2 (29 m/s)^2
313.6 = 1/2 (841)
313.6 = 420.5
To correct the mistake,
The equation should be:
gh + 0 = 1/2 v^2
Simplifying further,
9.8 m/s^2 * 32 m = 1/2 * (29 m/s)^2
313.6 = 1/2 * 841
313.6 = 420.5/2
313.6 = 210.25
Now, to solve for i, multiply both sides of the equation by 2:
2 * 313.6 = 2 * 210.25
627.2 = 420.5
Therefore, the initial speed (i) of the rock is approximately 627.2 m/s.
Now, let's move on to finding the greatest height of the rock (ii).
The greatest height of the rock can be found using the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
where vf is the final velocity (which is 0 m/s at the highest point), vi is the initial velocity (which is the value we are looking for), a is the acceleration due to gravity (-9.8 m/s^2), and d is the displacement (which is the greatest height).
Plugging in the values:
0^2 = (627.2 m/s)^2 + 2(-9.8 m/s^2)d
0 = 394020.64 m^2/s^2 - 19.6 m/s^2 d
19.6 m/s^2 d = 394020.64 m^2/s^2
d = 394020.64 m^2/s^2 / 19.6 m/s^2
d = 20100.53 m
Therefore, the greatest height (ii) of the rock, as measured from the base of the cliff, is approximately 20,100.53 m.