Question

A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope without rolling). The slope makes an angle θ with the horizontal. Using conservation of energy, find the maximum vertical height hmax to which the ball will climb.

Answer 1

Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving  for hmax:

hmax = 1/2 · v² / g

Answer 2

The maximum vertical height hmax to which the ball will climb by the law of conservation of energy is,

$h_{max}=\dfrac{1}{2g}v^2\cos\theta$

What is conservation of energy?

According to the law of conservation of energy, the total amount of energy of an isolated system remain constant.

For the conservation of energy, the sum of initial kinetic energy and the potential energy of the system is equal to the sum of final kinetic energy and the potential energy of the system.

It can be given as,

$K_i+U_i=K_f+U_f$

Here, (K) represent the kinetic energy and (U) represent the potential energy of the system.

A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope without rolling). The initial kinetic energy of the ball is given as,

$K_i=\dfrac{1}{2}mv^2$

The slope makes an angle θ with the horizontal. Here, the velocity in the y-direction will be equal to zero and velocity in x-direction is $v\cos\theta$. Therefore,

$K_i=\dfrac{1}{2}mv^2\cos\theta$

The potential energy at maximum height, (h{max}) is,

$U_f=mgh_{max}$

At the maximum height, the final potential energy which ball posses will be equal to the intial kinetic energy of it. Thus by the conservation of energy,

$mgh_{max}=\dfrac{1}{2}mv^2\cos\theta\\h_{max}=\dfrac{1}{2g}v^2\cos\theta$

Hence, the maximum vertical height hmax to which the ball will climb by the law of conservation of energy is,

$h_{max}=\dfrac{1}{2g}v^2\cos\theta$

Learn more about the conservation of energy here;

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