A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½ at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):
v^2 = u^2 + 2as
v^2 = 17.0^2 -2(9.81)(11.0)
v = √73.18 = 8.55m/s
now from equation (3):
v = u + at
8.55 = 17.0 – 9.81t
t = (8.55 – 17.0)/(-9.81) = 0.86s
Given that,
Mass of object, m = 55 kg
Mechanical energy of the object, M = 4306 J
Potential energy, P = 2940 J
We know that the mechanical energy is the sum of kinetic and potential energy such that,
Mechanical energy = kinetic energy + potential energy
Kinetic energy is given by :
v is velocity of object
So, the velocity of object is 7.04 m/s.
I have all the answers here so take this
