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.

Photo AI

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 $g$

96%

114 rated

Answer

To find the time $T$ at which the particle reaches its maximum height, we use the first equation of motion, which states that the final velocity $v = u + at$ , where $a$ is the acceleration due to gravity (which is negative in this case).

At maximum height, the final velocity $v = 0$ . Therefore:

$0 = u - gT$

Rearranging gives:

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

Step 2

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

99%

104 rated

Answer

We can find the maximum height $H$ using the second equation of motion:

$H = uT + \frac{1}{2}aT^2$

Substituting $T = \frac{u}{g}$ and $a = -g$ yields:

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

Simplifying gives:

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

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

Step 3

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

96%

101 rated

Answer

The height of point $A$ is $3H$ , which can be substituted as:

$A = 3H = 3 \cdot \frac{u^2}{2g} = \frac{3u^2}{2g}$

To find the total time when the particle hits the ground, we can use the fact that the time to go up to the maximum height is equal to the time to come down. Therefore:

The time of ascent is $T$ , and the total time of descent can be found using the total distance:

Using the equation:

$s = ut + \frac{1}{2}at^2$

Here, we set up the equation for total vertical distance:

$-\frac{3u^2}{2g} = 0 + \frac{1}{2}(-g)t^2$

Solving for $t$ gives:

$t^2 = \frac{3u^2}{g} \\ t = \sqrt{\frac{3u^2}{g}}$

Total time from projection to hitting the ground is:

$T_{total} = T + t = T + 2T = 3T$

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

Join the A-Level students using SimpleStudy...