Answer to Question #161897 in Mechanics | Relativity for Mercy

Question

A stone is projected at an angle 60 degrees to the horizontal with the velocity of 30m/s. Calculate the highest point reached, the range, the time taken for flight and the height of the stone at the instant that the path makes an angle 30 degrees with the horizontal

Answer 1

a) Let's first find the time that projectile takes to reach the highest point:

$$v_y=v_0sin\theta-gt,$$ $$0=v_0sin\theta-gt,$$ $$t=\dfrac{v_0sin\theta}{g}.$$

The equation for the motion of projectile in the vertical direction can be written as follows:

$$y=v_0sin\theta-\dfrac{1}{2}gt^2.$$

Substituting $t$ into the equation we can find the highest point reached by the projectile:

$$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(30\ \dfrac{m}{s})^2sin^260^{\circ}}{2\cdot9.8\ \dfrac{m}{s^2}}=34.44\ m.$$

b)-c) Let's first find the time taken for flight:

$$t_{flight}=2t=\dfrac{2v_0sin\theta}{g},$$ $$t_{flight}=\dfrac{2\cdot30\ \dfrac{m}{s}\cdot sin60^{\circ}}{9.8\ \dfrac{m}{s^2}}=5.3\ s.$$

Then, we can find the range of the projectile:

$$x=v_0t_{flight}cos\theta,$$ $$x=30\ \dfrac{m}{s}\cdot5.3\ s\cdot cos60^{\circ}=79.5\ m.$$

(d) At the point when the path makes an angle of $30^{\circ}$ to the horizontal we have:

$$tan30^{\circ}=\dfrac{v_y}{v_x}=\dfrac{v_0^2sin^2\theta-2gh}{v_0cos\theta},$$ $$h=\dfrac{v_0^2sin^2\theta-v_0cos\theta tan30^{\circ}}{2g},$$ $$h=\dfrac{(30\ \dfrac{m}{s})^2sin^260^{\circ}-30\ \dfrac{m}{s}\cdot cos60^{\circ}\cdot tan30^{\circ}}{2\cdot9.8\ \dfrac{m}{s^2}}=34\ m.$$