At time $t = 0$, a particle is projected vertically upwards with speed $u$ from a point $A$. The particle moves freely under gravity. At time $T$ the particle is at its maximum height $H$ above $A$. (a) Find $T$ in terms of $u$ and $g$. (b) Show that $H = \frac{u^2}{2g}$. The point $A$ is at a height $3H$ above the ground. (c) Find, in terms of $T$, the total time from the instant of projection to the instant when the particle hits the ground.

A-Level Edexcel Maths Mechanics Forces: At time $t = 0$, a particle is p

At time $t = 0$, a particle is projected vertically upwards with speed $u$ from a point $A$ - Edexcel - A-Level Maths Mechanics - Question 4 - 2014 - Paper 1

Question 4

At time $t = 0$, a particle is projected vertically upwards with speed $u$ from a point $A$. The particle moves freely under gravity. At time $T$ the particle is at ... show full transcript

Worked Solution & Example Answer:At time $t = 0$, a particle is projected vertically upwards with speed $u$ from a point $A$ - Edexcel - A-Level Maths Mechanics - Question 4 - 2014 - Paper 1

Step 1

Find $T$ in terms of $u$ and $g$

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Answer

To find the time $T$ it takes for the particle to reach its maximum height, we use the kinematic equation for motion under uniform acceleration:

$v = u - gT$

At the maximum height, the final velocity $v = 0$ . Therefore, we can set up the equation:

$0 = u - gT$

Rearranging the equation gives:

$gT = u$

Thus,

$T = \frac{u}{g}.$

Step 2

Show that $H = \frac{u^2}{2g}$

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Answer

The maximum height $H$ of the particle can be calculated using the following kinematic equation:

$H = ut + \frac{1}{2} a t^2,$

where $a = -g$ (acceleration due to gravity) and $t = T$ .

Substituting for $T$ gives:

$H = uT + \frac{1}{2} (-g) T^2$

Replacing $T$ with $\frac{u}{g}$ results in:

$H = u \left( \frac{u}{g} \right) - \frac{1}{2} g \left( \frac{u}{g} \right)^2$

Simplifying this expression leads to:

$H = \frac{u^2}{g} - \frac{1}{2} \frac{u^2}{g} = \frac{u^2}{2g}.$

Step 3

Find, in terms of $T$, the total time from the instant of projection to the instant when the particle hits the ground

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Answer

The total time taken for the particle to hit the ground after reaching the maximum height can be found by considering the symmetry of projectile motion.

The time to reach maximum height is $T = \frac{u}{g}$ .
Since the time to go up is equal to the time to come down, the total time $T_{total}$ is:

$T_{total} = 2T = 2 \left( \frac{u}{g} \right) = \frac{2u}{g}.$

At time $t = 0$, a particle is projected vertically upwards with speed $u$ from a point $A$ - Edexcel - A-Level Maths Mechanics - Question 4 - 2014 - Paper 1

Worked Solution & Example Answer:At time $t = 0$, a particle is projected vertically upwards with speed $u$ from a point $A$ - Edexcel - A-Level Maths Mechanics - Question 4 - 2014 - Paper 1

Find $T$ in terms of $u$ and $g$

Show that $H = \frac{u^2}{2g}$

Find, in terms of $T$, the total time from the instant of projection to the instant when the particle hits the ground

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