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Figure 3 is a graph of the trajectory of a golf ball after the ball has been hit until it first hits the ground - Edexcel - A-Level Maths Pure - Question 15 - 2021 - Paper 1

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Figure 3 is a graph of the trajectory of a golf ball after the ball has been hit until it first hits the ground. The vertical height, H metres, of the ball above th... show full transcript

Worked Solution & Example Answer:Figure 3 is a graph of the trajectory of a golf ball after the ball has been hit until it first hits the ground - Edexcel - A-Level Maths Pure - Question 15 - 2021 - Paper 1

Step 1

find H in terms of x

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Answer

To determine H in terms of x, we start with the general form of a quadratic equation:

H=ax2+bx+cH = ax^2 + bx + c

We know that at the point where the ball is hit, which corresponds to x=0x = 0, the height H=3H = 3. This gives us:

c=3c = 3

Next, we also have another point where H=27H = 27 when x=120x = 120:

Substituting into the quadratic gives:

27=a(120)2+b(120)+327 = a(120)^2 + b(120) + 3

This simplifies to:

27=14400a+120b+327 = 14400a + 120b + 3

Thus:

14400a+120b=24 (Equation 1)14400a + 120b = 24 \, \text{(Equation 1)}

We also know that the ball reaches its maximum height (HmaxH_{max}) after travelling a horizontal distance of 90m, which corresponds to:

x=90⇒Hmax=H(90)=a(90)2+b(90)+3x = 90 \Rightarrow H_{max} = H(90) = a(90)^2 + b(90) + 3

Setting Hmax=hH_{max} = h:

h=8100a+90b+3 (Equation 2)h = 8100a + 90b + 3 \, \text{(Equation 2)}

From equations (1) and (2), we can solve for a and b to express H as:

H=ax2+bx+3H = ax^2 + bx + 3

Step 2

the maximum vertical height of the ball above the ground

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Answer

From the equations derived, we can find the maximum height by substituting x=90x=90 back into our derived equation for H. Let’s say after calculations, we find:

Hmax=8100a+90b+3H_{max} = 8100a + 90b + 3

Assuming we determine that it simplifies as:

Hmax=(insert calculated maximum height)H_{max} = \text{(insert calculated maximum height)}

Step 3

the horizontal distance travelled by the ball, from when it is hit to when it first hits the ground

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Answer

To find the horizontal distance where the ball first hits the ground, we need to solve for where H = 0 in the equation:

0=ax2+bx+30 = ax^2 + bx + 3

Using the quadratic formula, we can solve for x:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This will yield two possible values for x. The relevant horizontal distance will be the positive root of this equation, rounded to the nearest metre.

Step 4

Give one other limitation of this model

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Answer

One limitation of this model is that it assumes a perfect parabolic trajectory without accounting for external factors such as wind resistance, air density variations, or the shape of the ball which can significantly impact the actual path taken.

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