# A golfer imparts a speed of 30.3 m/s to a ball, and it travels the max

A golfer imparts a speed of 30.3 m/s to a ball, and it travels the maximum possible distance before landing on the green.
The tee and the green are at the same elevation.
(a) How much time does the ball spend in the air?
(b) What is the longest “hole in one" that the golfer can make, if the ball does not roll when it hits the green?
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Becky Harrison
(a) The ball will travel the maximum distance if the it is launched at an angle of $\theta ={45}^{\circ }$, to find the time of flight we can use the following equation in the vertical plane,
$y={v}_{0y}t+\frac{1}{2}a{t}^{2}$
where y is the displacement when the ball lands on the green, so it equals zero, since the ball starts and ends at the same elevation. And the initial velocity in the vertical component is given with ${v}_{0y}={v}_{0}\mathrm{sin}\left(\theta \right)$, hence:
$0={v}_{0}\mathrm{sin}\left(\theta \right\}t+\frac{1}{2}a{t}^{2}$
$\frac{1}{2}at=-{v}_{0}\mathrm{sin}\left(\theta \right)$
$t=-\frac{2{v}_{0}\mathrm{sin}\left(\theta \right)}{a}$
substitute with the given values to get:
$t=-\frac{2\left(30m{×}^{-1}\right)\mathrm{sin}\left(45\right)}{\left(-9.8m{×}^{-2}\right)}$
$=4.33s$
$t=4.33s$
(b) The horezantal displacemnet can be found from:
$x={v}_{0x}t+\frac{1}{2}a{t}^{2}$
where ${v}_{0x}={v}_{0x}\mathrm{cos}\left(\theta \right)$ is the initial velocity in the vertical component, and a is the acceleration in the horizontal component which equals zero, so:
$x={v}_{0}t\mathrm{cos}\left(\theta \right)$
substitute with the given values to get:
$\mid x=\left(30m×{s}^{-1}\right)\left(4.33m\right)\mathrm{cos}\left(45\right)$
$=91.85m$
$x=91.85m$

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Thomas Lynn

Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, $v=u+at$, where vis the final velocity, u is.
the initial velocity, a is the acceleration and tis the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity $=u\mathrm{sin}\theta$, acceleration = acceleration due to gravity $=-{g}^{m\wedge S}$ and final velocity = 0 m/s.
$O=u\mathrm{sin}\theta -gt$
$t=u\frac{\mathrm{sin}\theta }{g}$
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile $=2u\frac{\mathrm{sin}\theta }{g}$
Horizontal motion:
We have equation of motion , $=ut+\frac{1}{2}a{t}^{2}$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = $u\mathrm{cos}\theta$, acceleration $=0m{s}^{2}$ and time taken $=2u\frac{\mathrm{sin}\theta }{g}$
So range of projectile,$R=u\mathrm{cos}\theta \star \frac{2u\mathrm{sin}\theta }{g}=\frac{{u}^{2}\mathrm{sin}2\theta }{g}$
a) We have golf ball travels maximum distance, so range is maximum.
Maximum range is when, $\mathrm{sin}2\theta =1$

Now we have travel time of projectile, $t=2u\frac{\mathrm{sin}\theta }{g}$
Initial velocity $=30.3\frac{m}{s}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\theta =45°$
So time spend in air,$t=\frac{2\star 30.3\star \mathrm{sin}45}{9.81}=4.368$ seconds
b) Longest hole that golfer can make = Range of projectile $=\frac{{u}^{2}\mathrm{sin}2\theta }{g}$
Longest hole that golfer can make $=\frac{{30.3}^{2}\mathrm{sin}\left(2\star 45\right)}{9.81}=93-59$meter

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