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(C) For the first projectile,the height at time $t$ is $h_1 = ut - \frac{1}{2}gt^2$.For the second projectile,which is fired after time $T$,the height

Vedclass · Accepted Answer

They meet at time $t = \frac{u}{g} + \frac{T}{2}$ and at a height $\frac{u^2}{2g} - \frac{gT^2}{8}$

Vedclass · Answer

(C) For the first projectile,the height at time $t$ is $h_1 = ut - \frac{1}{2}gt^2$.For the second projectile,which is fired after time $T$,the height at time $t$ is $h_2 = u(t - T) - \frac{1}{2}g(t - T)^2$.When both meet,$h_1 = h_2$.$ut - \frac{1}{2}gt^2 = u(t - T) - \frac{1}{2}g(t - T)^2$.Expanding the right side: $ut - \frac{1}{2}gt^2 = ut - uT - \frac{1}{2}g(t^2 - 2tT + T^2)$.$ut - \frac{1}{2}gt^2 = ut - uT - \frac{1}{2}gt^2 + gtT - \frac{1}{2}gT^2$.Canceling common terms: $0 = -uT + gtT - \frac{1}{2}gT^2$.$uT + \frac{1}{2}gT^2 = gtT$.Dividing by $gT$: $t = \frac{u}{g} + \frac{T}{2}$.Substituting $t$ back into $h_1$: $h_1 = u(\frac{u}{g} + \frac{T}{2}) - \frac{1}{2}g(\frac{u}{g} + \frac{T}{2})^2$.$h_1 = \frac{u^2}{g} + \frac{uT}{2} - \frac{1}{2}g(\frac{u^2}{g^2} + \frac{uT}{g} + \frac{T^2}{4})$.$h_1 = \frac{u^2}{g} + \frac{uT}{2} - \frac{u^2}{2g} - \frac{uT}{2} - \frac{gT^2}{8}$.$h_1 = \frac{u^2}{2g} - \frac{gT^2}{8}$.

$A$ projectile is fired vertically upwards with an initial velocity $u$. After an interval of $T$ seconds,a second projectile is fired vertically upwards,also with initial velocity $u$.

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