Physics, 01.11.2020 09:20 mobslayer88

A golfer hits a shot to an elevated green. The ball leaves the club with an initial speed of 16 m/s at an angle of 58° above the hori-
zontal. If the speed of the ball just before it lands is 12 m/s, what is
the elevation of the green above the point where the ball is struck?

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Answer from: moongirl4791

Approximately $5.7\; \rm m$ , assuming that the air resistance on this golf ball is negligible.

Explanation:

The mechanical energy of an object is the sum of its potential and kinetic energy.

If the air resistance on this golf ball is negligible, the mechanical energy of this golf ball should stay the same while it was in the air.

However, the final height of this ball is greater than its initial height. The ball has gained (gravitational) potential energy during the flight.

The mechanical energy of this ball has to stay the same during the flight. There was no energy input to the ball other than the initial push. Therefore, every joule of the potential energy that the ball has gained should correspond to one joule of kinetic energy that the ball loses.

The difference between the initial and final kinetic energy of this ball will thus be equal to the potential energy that the ball has gained. The elevation that the ball has gained may be found from the increase in the potential energy of this ball.

Steps

Let denote the mass of this golf ball.

Let v_0 and v_1 denote the initial and final (right-before-landing) speed of this ball, respectively. (The question states that $v_0 = 16\; \rm m \cdot s^{-1}$ while $v_0 = 12\; \rm m \cdot s^{-1}$ .)

Let denote the gravitational field strength. ( $g \approx 9.81\; \rm m\cdot s^{-1}$ .)

The initial kinetic energy of this golf ball would be:

$\displaystyle \text{Initial KE} = \frac{1}{2}\, m \cdot {v_0}^2$ .

The final (right-before-landing) kinetic energy of this golf ball would be:

$\displaystyle \text{Final KE} = \frac{1}{2}\, m \cdot {v_1}^2$ .

$\text{Initial KE} \text{Final KE}$ because v_0 < v_1 and m 0 .

Calculate the kinetic energy that this golf ball has lost during the flight:

$\begin{aligned}& \text{KE Lost} \\ &= \text{Initial KE} - \text{Final KE} \\ &= \frac{1}{2}\, m\cdot {v_0}^2 - \frac{1}{2}\, m\cdot {v_1}^2 = \frac{1}{2}\, m \,\left( {v_0}^2 - {v_1}^2\right)\end{aligned}$ .

Based on the assumptions, the kinetic energy that this golf ball has lost should be equal (in size) to the potential energy that this golf ball has gained:

$\begin{aligned} \text{PE Gained} = \text{KE Lost} &= \frac{1}{2}\, m \,\left( {v_0}^2 - {v_1}^2\right)\end{aligned}$ .

The potential energy that this golf ball has gained is proportional to the increase in its height. Let $\Delta h$ denote the elevation that this golf ball has gained.

$\text{PE Gained} = m \cdot g \cdot \Delta h$ .

Rearrange this equation and solve for $\Delta h$ :

$\begin{aligned}&\Delta h \\ &= \frac{\text{PE Gained}}{m \cdot g} \\ &= \frac{(1/2)\, m \,\left( {v_0}^2 - {v_1}^2\right)}{m \cdot g} = \frac{{v_0}^2 - {v_1}^2}{2\, g}\end{aligned}$ .

Given that $v_0 = 16\; \rm m \cdot s^{-1}$ , $v_1 = 12\; \rm m \cdot s^{-1}$ , and assume that $g \approx 9.81\; \rm m \cdot s^{-2}$ , calculate the elevation that this golf ball has gained:

$\begin{aligned}&\Delta h \\ &= \frac{{v_0}^2 - {v_1}^2}{2\, g} \\ &= \frac{\left(16\; \rm m \cdot s^{-1}\right) - \left(12\; \rm m \cdot s^{-1}\right)}{2\times 9.81\; \rm m \cdot s^{-2}} \approx 5.71\; \rm m\end{aligned}$ .

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A golfer hits a shot to an elevated green. The ball leaves the club with an initial speed of 16 m/s...

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