A golf ball with an initial angle of 34 degrees lands exactly 240 m down the range on a level course.
a.) Neglecting air friction, what initial speed would achieve this result?
b.) Using the speed determined in item a, find the maximum height reached by the ball.
I was looking up solutions online and I stumbled across this explanation:
Let the initial speed be v
Initial vertical speed is v * sin(34)
Time of the flight of the ball will be [2v * sin(34)]/g
Initial horizontal speed is v * cos(34)
Then, they did the following:
The range = [2v * sin(34)]/g * (v * cos(34)
= v^2 sin * 2 * 34 / g
= v^2 sin(68)/g
My question is how they got two things. 1) The formula for the time of the flight of the ball. Second. When they multiplied the the range function, they somehow eliminated the cos and were left with only a sin? Any help is appreciated!!!
To answer your questions:
1) The formula for the time of flight of the ball is derived by considering the vertical motion of the ball. Since the initial vertical speed is v * sin(34), the ball will reach the ground when its vertical displacement is equal to zero. Using the equation of motion for vertical motion:
0 = (v * sin(34)) * t - (1/2) * g * t^2
where t is the time of flight and g is the acceleration due to gravity. Rearranging the equation and solving for t will give you the formula [2v * sin(34)]/g.
2) When multiplying the range function, which is [2v * sin(34)]/g * (v * cos(34)), you can distribute the terms:
[2v * sin(34)]/g * (v * cos(34)) = (2v * sin(34) * v * cos(34))/g
Then, using the trigonometric identity sin(2θ) = 2sin(θ)cos(θ), you can rewrite the expression as:
(2v * sin(34) * v * cos(34))/g = v^2 * sin(68)/g
So, by applying the trigonometric identity, they eliminated the cos term and were left with only the sin term.
To find the formula for the time of flight of the ball, the online solution used the vertical motion equation:
\[S = ut + \frac{1}{2}gt^2\]
Where:
S = displacement in the y-direction (equal to 0 because the ball lands at the same level it was launched)
u = initial vertical velocity (v * sin(34))
t = time of flight
g = acceleration due to gravity
Since the displacement in the y-direction is 0, the equation becomes:
\[0 = (v * \sin(34))t + \frac{1}{2}gt^2\]
\[0 = (v * \sin(34))t + \frac{1}{2}(9.8)t^2\]
Dividing both sides of the equation by t, we get:
\[0 = (v * \sin(34)) + \frac{1}{2}(9.8)t\]
Rearranging the equation, we have:
\[(v * \sin(34)) = -\frac{1}{2}(9.8)t\]
By substituting the value of t from the equation above into the range equation, the solution was able to eliminate the cos component of the range equation.
Now, let's calculate the range equation step by step to understand how they eliminated the cos term:
Given that the range R = 240 m, we can use the horizontal motion equation:
\[R = u \cdot t\]
Where:
R = range (240 m)
u = initial horizontal velocity (v * cos(34))
t = time of flight
Substituting the values into the equation:
\[240 = (v * \cos(34)) \cdot t\]
Rearranging the equation:
\[t = \frac{240}{(v * \cos(34))}\]
Substituting the value of t into the range equation, we get:
\[R = v \cdot \sin(34) \cdot \frac{240}{(v * \cos(34))}\]
Cancelling out the v in the numerator and denominator:
\[R = \frac{240 \cdot \sin(34)}{\cos(34)}\]
Using the identity \(\tan(x) = \frac{{\sin(x)}}{{\cos(x)}}\), we can simplify the equation:
\[R = 240 \cdot \tan(34)\]
So the range equation becomes:
\[R = v^2 \cdot \frac{{\sin(68)}}{{g}}\]
By substituting the value of \(v^2\sin(68)/g\) into the range equation, the online solution eliminated the cos term.
time in air comes from
hf=hi-vvertical*t-1/2 g t^2
but hf=hi=zero so you can solve for time in air.
Vvertical=v*sinTheta
so t=2VsinTheta/g
horizonal distance:
distance=vhorizontal*timeinair
= v costheta*t
so you have a final distance which has a cosTheta*sinTheta
in it. do you remember the double angle formula
Sin(2Theta)=cosTheta*sinTheta