A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 27 m high. When it hits the ground at the base of the cliff the rock has a speed of 24 m/s.
(a) Assuming that air resistance can be ignored, find the initial speed of the rock.
(b) Find the greatest height of the rock as measured from the base of the cliff.
I know I have to use projectile motion somehow. I don't know how to without knowing the initial velocity. I can't find that without solving the first part. I'm having trouble picking an equation. Please help.
Initial KE+ initialPE=final KE
I suspect you can solve it from that.
Greatest height? Final KE= greatest PE
solve for greatest height.
KE initial = 0
PE initial = mgh = 68.8662 = KE final
KE final = 1/2mv2
v = 23.01
That isn't right. What have I done wrong?
Where did you get initial KE is zero? That is what you are solving for. You need a tutor, pronto.
I am just so confused by this problem. I have done other problems with ease and I'm having so much trouble with other ones.
you do:
1/2mv(initial)^2+mgh=1/2mv(final)^2
so simplify that to : v(inital)= sqrt(v(final)^2-2gh)
For b): 1/2mv(initial)^2=mgh
solve for h : h=v^2/2g
then add your original height to the new height for your answer
To solve this problem, you can use a combination of the kinematic equations for projectile motion. Let's break it down step by step:
(a) To find the initial speed (also known as the initial velocity) of the rock, you need to determine the speed at the instant it was thrown vertically upward from the top of the cliff. This is when its speed is zero.
You can use the equation for vertical motion:
v^2 = u^2 + 2as
where:
v = final velocity (24 m/s in this case, as given)
u = initial velocity (what we need to find)
a = acceleration (acceleration due to gravity, which is approximately -9.8 m/s^2)
s = displacement (vertical height of the cliff, which is 27 m)
Plugging in the given values, the equation becomes:
0 = u^2 + 2(-9.8)(27)
Simplifying and solving for u^2 will give you the square of the initial velocity. Remember that the initial velocity itself can be positive or negative depending on the direction of motion. In this case, since the rock is thrown vertically upward, the initial velocity is negative because it is against the direction of gravity.
(b) To find the greatest height reached by the rock, you need to find the displacement covered by the rock from the top of the cliff to the highest point of its trajectory.
You can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity (zero at the highest point)
u = initial velocity (found in part (a))
a = acceleration (acceleration due to gravity, approximately -9.8 m/s^2)
s = displacement (what we need to find)
Plugging in the given values, the equation becomes:
0 = u^2 + 2(-9.8)s
Now, solve this equation for s to find the greatest height reached by the rock from the base of the cliff.
By using these equations and the given values, you can determine the initial speed of the rock and the greatest height it reaches.