Question

A body is to be projected from the ground to have a range of 45m.If the horizontal component of its initial velocity is 15m/s ,the angle of projection is where g=10

Answer 1

Let's first find the total time of flight of the body:

$$x=v_{0x}t_{flight},$$ $$t_{flight}=\dfrac{x}{v_{0x}}=\dfrac{45\ m}{15\ \dfrac{m}{s}}=3\ s.$$

Let's find the time that the body takes to reach its maximum height:

$$v_y=v_{0y}-gt,$$ $$0=v_{0y}-gt,$$ $$t=\dfrac{v_{0y}}{g}.$$

Since, $t_{flight}=2t$ we can write:

$$\dfrac{t_{flight}}{2}=\dfrac{v_{0y}}{g}.$$

From this equation we can find the vertical component of body's initial velocity:

$$v_{0y}=\dfrac{1}{2}t_{flight}g=\dfrac{1}{2}\cdot3\ s\cdot10\ \dfrac{m}{s^2}=15\ \dfrac{m}{s}.$$

Let's write horizontal and vertical components of body's initial velocity:

$$v_{0x}=v_0cos\theta, (1)$$ $$v_{0y}=v_0sin\theta. (2)$$

Dividing the second equation by the first one, we can find the angle of projection:

$$tan\theta=\dfrac{v_{0y}}{v_{0x}}=\dfrac{15\ \dfrac{m}{s}}{15\ \dfrac{m}{s}}=1,$$ $$\theta=tan^{-1}(1)=45^{\circ}.$$