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Two particles $P$ and $Q$ are projected vertically upwards from horizontal ground at the same instant.

The speeds of projection of $P$ and $Q$ are $12{\text{ }}m{\text{ }}{s^{ - 1}}$ and $7{\text{ }}m{\text{ }}{s^{ - 1}}$ respectively and the heights of $P$ and $Q$ above the ground, $t$ seconds after projection, are ${h_P}m$ and ${h_Q}m$ respectively. Each particle comes to rest on returning to the ground.

a) Find the set of values of $t$ for which the particles are travelling in opposite directions.

b) At a certain instant, $P$ and $Q$ are above the ground and $3{h_P} = 8{h_Q}$. Find the velocities of $P$ and $Q$ at this instant.

پاسخ تشریحی :
نمایش پاسخ

a) Times to max. height are $1.2s$ and $0.7s$

Range of values is ${\text{0}}{\text{.7}} \lt {\text{t}} \lt {\text{1}}{\text{.12}}$

b) $36t - 1.5g{t^2} = 56t - 4g{t^2}$

$t = 8/g$

Alternative for part 5 (b)

For using $3{h_P} = 8{h_Q} \to 3\left( {{v_P}^2 - 144} \right) \div $

$\left( { - 20} \right) = 8\left( {{v_Q}^2 - 49} \right) \div \left( { - 20} \right) \to 3{v_P}^2 - 8{v_Q}^2$

$ = 40$

For using ${v_P} = 12 - 10t$ and ${v_Q} = 7 - 10t$

$ \to {v_P} - {v_Q} = 5$

For eliminating ${v_Q}$ (or ${v_P}$) and solving for ${v_P}$ (or ${v_Q}$).

${v_P}^2 - 16{v_P} + 48 = 0 \to {v_P} = 4$ (or $4$, $12$)

Upward velocities are $4{\text{ }}m{s^{ - 1}}$ and $ - 1{\text{ }}m{s^{ - 1}}$

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