Question

If x component of velocity vix=4 m/s and y component of velocity viy=5 m/s then what is the range of projectile, height and time of flight ?

Answer 1

Let's first find the initial velocity of the projectile and the launch angle. We can find the initial velocity of the projectile from the Pythagorean theorem:

$$v_i=\sqrt{v_{ix}^2+v_{iy}^2}=\sqrt{(4\ \dfrac{m}{s})^2+(5\ \dfrac{m}{s})^2}=6.4\ \dfrac{m}{s}.$$

We can find the launch angle from the geometry:

$$\theta=tan^{-1}(\dfrac{v_{iy}}{v_{ix}}),$$ $$\theta=tan^{-1}(\dfrac{5\ \dfrac{m}{s}}{4\ \dfrac{m}{s}})=51.3^{\circ}.$$

(a) We can find the range of projectile as follows:

$$R=\dfrac{v_i^2sin2\theta}{g}=\dfrac{(6.4\ \dfrac{m}{s})^2\cdot sin2\cdot51.3^{\circ}}{9.8\ \dfrac{m}{s^2}}=4.07\ m.$$

(b) We can find the maximum height of the projectile as follows:

$$H=\dfrac{v_i^2sin^2\theta}{2g}=\dfrac{(6.4\ \dfrac{m}{s})^2\cdot sin^251.3^{\circ}}{2\cdot9.8\ \dfrac{m}{s^2}}=1.27\ m.$$

(c) We can find time of flight of the projectile as follows:

$$t=\dfrac{2v_isin\theta}{g}=\dfrac{2\cdot6.4\ \dfrac{m}{s}\cdot sin51.3^{\circ}}{9.8\ \dfrac{m}{s^2}}=1.02\ s.$$