Question

A projectile is fired with an initial speed ${\mathit{v}}_{\mathbf{0}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$from level ground at a target that is on the ground, at distance $\mathit{R}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$, as shown in Fig. 4-59. What are the (a) least and (b) greatest launch angles that will allow the projectile to hit the target?

Short answer

(a)Least angle of projection is$6.{29}^0$

(b)Greatest angle of projection is$83.7^0$

Step-by-step solution

Given information

It is given that,

$v_0=30.0\mathrm{m}/\mathrm{s}$

$R=20.0\mathrm{m}$

$a=9.8\mathrm{m}/\mathrm{s}$

Determining the concept

This problem involves kinematic equation of motion that describe the motion of an object with constant acceleration. Also it deals with the resolution of vectors. The resolution of a vector is the splitting of a single vector into two or more vectors in different directions. Here, the angle of projection can be found using components of initial velocity and the distance. The times will be different for different angles. Using kinematic equations, find the angles.

Formula is as follow:

$d=v_{0}t+\left(\frac{1}{2}\right)a{t}^2$$d=v_{0}t+\left(\frac{1}{2}\right)a{t}^2$ (i)

where, dis total displacements, $v_0$ is an initial velocity, t is time and a is an acceleration.

(a) Determining the least angle of projection

Now, resolve velocity in x and y components,

$\begin{array}{ll}{v}_{0x}=v_0& \cos \theta \\ v_{0y}=v_0& \sin \theta \end{array}$

We use the formula of distance for vertical and horizontal range using respective components of initial velocities. The time of flight will be the same for both ranges.

$d_x=v_{0x}t$

$R=v_{0x}t$

localid="1657022089885" $20.0=30.0\cos \left(\theta \right)t$

$t=\frac{20.0}{30.0\cos \theta }$

$d_y=V_{0y}t+\left(\frac{1}{2}\right)a{t}^2$

$0=30.0\sin \left(\theta \right)t+\left(\frac{1}{2}\right)\left(-9.8\right)t^2$

Plug value of t from above,

$0=30.0\sin \left(\theta \right)\frac{20.0}{30.0\cos \left(\theta \right)}-\left(\frac{1}{2}\right)(9.8){\left(\frac{20.0}{30.0\cos \left(\theta \right)}\right)}^2$

$4.9{\left(\frac{20.0}{30.0\cos \left(\theta \right)}\right)}^2=30.0\sin \left(\theta \right)\frac{20.0}{30.0\cos \left(\theta \right)}$

$4.9\left(\frac{20.0}{30.0\cos \left(\theta \right)}\right)=(30.0\sin \left(\theta \right)$

$$\begin{gathered} 98=30.0\sin \left(\theta \right)30.0\cos \left(\theta \right) \\ 98=900\sin \left(\theta \right)\cos \left(\theta \right) \end{gathered}$$

$\frac{98}{900}=\frac{\sin \left(2\theta \right)}{2}$

$\sin \left(2\theta \right)=2\left(\frac{98}{900}\right)$

localid="1657021544650" $=0.2177$

$2\theta =\left(0.2177\right)$

localid="1657022210522" $\theta =\frac{\left(0.2177\right)}{2}=6.{29}^0$

Therefore, the least angle of projection islocalid="1657022226831" $6.{29}^0$

(b) Determining the greatest angle of projection

Now, the angle should not be more than ${90}^0$so the greatest angle of projection should be,

localid="1657022270466" $$\begin{gathered} \theta =90-6.{29}^0 \\ =83.7^0 \end{gathered}$$

Therefore, the greatest angle of projection is$83.7^0$.