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.

A-Level Edexcel Maths Mechanics Projectiles: A ball is thrown vertically upwa

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

Question 2

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

96%

114 rated

Answer

To find the initial speed, we can use the kinematic equation which relates initial velocity, final velocity, acceleration, and time:

$v = u + at$

Here, the final speed $v = -6.45$ m/s (downwards), acceleration $a = -9.8$ m/s² (due to gravity), and time $t = 0.75$ s.

Using the equation:

$-6.45 = u - 9.8 \times 0.75$

Rearranging the equation gives:

$u = -6.45 + 9.8 \times 0.75$

Calculating this:

$u = -6.45 + 7.35 = 0.9$

Step 2

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

99%

104 rated

Answer

Using the kinematic equation again with the initial speed found, we set the final speed at the highest point as 0:

$0 = 0.81 - 2 \times 9.8 \times s$

Rearranging gives:

$s = \frac{0.81}{2 \times 9.8} \approx 0.041 \text{ or } 0.0413 ext{ m}$

Step 3

Find the value of h.

96%

101 rated

Answer

To find the value of h, we can use the following kinematic equation:

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

Calculating this gives:

$h = -0.675 + 4.9 \times 0.5625 = -0.675 + 2.765625$

Thus,

$h \approx 2.1 \text{ or } 2.08 ext{ m}$

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.

Join the A-Level students using SimpleStudy...