Question

A bullet is fired with initial velocity of 100 m/s from the hill at point P at an angle of 30° with the horizontal and it strikes at 80 m lower than P. Neglecting air resistance and by using the concept of linear and curvilinear motion, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground, (b) the greatest elevation above the ground reached by the projectile.

Answer 1

$v_x = 100\cos30° = 86.6\ m/s\\ v_y = 100\sin 30° = 50\ m/s$

$H = \dfrac{u²\sin²\theta}{2g}= \dfrac{100²×0.25}{2×9.8} =$

$127.55m$

$H_{max} =80 + 127.55 =207.55m$

$y =u_yt + 0.5at²\\ 207.55 =50t + 0.5(9.8)t²\\ 4.9t²+ 50t-207.55= 0$

$t = 3.16s$

$\begin{aligned} a. \ \ x &= u_xt\\ x &= 86.6(3.16)= 273.656m\end{aligned}$

$b. \ \ H_{max } = 207.55m$