A ball is thrown vertically upwards with speed u m s⁻¹ from a point P at height h metres above the ground. The ball hits the ground 0.75 s later. The speed of the ball immediately before it hits the ground is 6.45 m s⁻¹. The ball is modelled as a particle. (a) Show that u = 0.9 (b) Find the height above P to which the ball rises before it starts to fall towards the ground again. (c) Find the value of h.

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A ball is thrown vertically upwards with speed u m s⁻¹ from a point P at height h metres above the ground - Edexcel - A-Level Maths Mechanics - Question 2 - 2011 - Paper 1

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A ball is thrown vertically upwards with speed u m s⁻¹ from a point P at height h metres above the ground. The ball hits the ground 0.75 s later. The speed of the ba... show full transcript

Worked Solution & Example Answer:A ball is thrown vertically upwards with speed u m s⁻¹ from a point P at height h metres above the ground - Edexcel - A-Level Maths Mechanics - Question 2 - 2011 - Paper 1

Step 1

Show that u = 0.9

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Answer

To find the initial speed u of the ball, we can use the kinematic equation which relates the final velocity (v), initial velocity (u), acceleration (a), and time (t):

$v = u + at$

Here:

v = -6.45 m/s (downward, hence negative)
a = -9.8 m/s² (acceleration due to gravity, negative as it acts downward)
t = 0.75 s

Substituting the values into the equation:

$-6.45 = u - 9.8 \times 0.75$

Calculating the right-hand side gives us:

$-6.45 = u - 7.35$

Rearranging to find u:

$u = -6.45 + 7.35$ $u = 0.9$

Hence, we have shown that u = 0.9.

Step 2

Find the height above P to which the ball rises before it starts to fall towards the ground again.

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Answer

When the ball rises to its maximum height, its final velocity v will be 0 m/s. We can again apply the kinematic equation:

$v^2 = u^2 + 2as$

Where:

v = 0 m/s
u = 0.9 m/s
a = -9.8 m/s² (acceleration due to gravity)
s = height above point P to which it rises.

Substituting the known values gives:

$0 = (0.9)^2 - 2 \times 9.8 \times s$

This simplifies to:

$0.81 = 19.6s$

Solving for s results in:

$s = \frac{0.81}{19.6} \approx 0.041 m$

Thus, the height above P is approximately 0.041 m.

Step 3

Find the value of h.

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Answer

We can now use the kinematic equations to find the total height h from which the ball was thrown. The total height can be calculated using:

$h = -0.9 \times 0.75 + 4.9 \times (0.75)^2$

Calculating this,

The term $-0.9 \times 0.75 = -0.675$ .
The term $4.9 \times (0.75)^2 = 4.9 \times 0.5625 = 2.75625$ .

Thus,

$h = -0.675 + 2.75625 = 2.08125$

Therefore, h is approximately 2.1 m or 2.08 m.

A ball is thrown vertically upwards with speed u m s⁻¹ from a point P at height h metres above the ground - Edexcel - A-Level Maths Mechanics - Question 2 - 2011 - Paper 1

Worked Solution & Example Answer:A ball is thrown vertically upwards with speed u m s⁻¹ from a point P at height h metres above the ground - Edexcel - A-Level Maths Mechanics - Question 2 - 2011 - Paper 1

Show that u = 0.9

Find the height above P to which the ball rises before it starts to fall towards the ground again.

Find the value of h.

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