Question

A ground to ground projectile is at point A at t $=\frac{T}{3}$, is at point B at $t=\frac{5T}{6}$ and reaches the ground at $t=T$. The difference in heights between points A and B is

• A. $\frac{g{T}^2}{6}$
• B. $\frac{g{T}^2}{12}$
• C. $\frac{g{T}^2}{18}$
• D. $\frac{g{T}^2}{24}$

Answer 1

The correct option is D $\frac{g{T}^2}{24}$

For ground to ground projectile,
$T=\frac{2u_y}{g}$
$\therefore u_y=\frac{gT}{2}$
Using equation of motion,
Height of projectile at A is
$h_A=u_y{t}_A-\frac{1}{2}g{t}_A^2$
$h_A=u_y\left(\frac{2u_y}{3g}\right)-\frac{1}{2}g{\left(\frac{2u_y}{3g}\right)}^2$
$h_A=\frac{4}{9}\frac{u_y^2}{g}=\left(\frac{4}{9g}\right){\left(\frac{gT}{2}\right)}^2=\frac{g{T}^2}{9}$
Height of projectile at B is
$h_B=u_y(\frac{5}{9}\times \frac{2u_y}{g})-\frac{1}{2}\times g\times {(\frac{5}{6}\times \frac{2u_y}{g})}^2$
$h_B=\frac{5}{18}\frac{u_y^2}{g}=\frac{5}{18g}{\left(\frac{gT}{2}\right)}^2=\frac{5}{72}g{T}^2$
$\therefore h_A-h_B=\frac{g{T}^2}{24}$.