91影视

It is given that,

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

$R=20.0\mathrm{m}$

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

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.

Now, resolve velocity in x and y components,

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

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" $20.0=30.0\cos \left(胃\right)t$

$t=\frac{20.0}{30.0\cos 胃}$

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

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

Plug value of t from above,

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

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

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

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

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

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

localid="1657021544650" $=0.2177$

$2胃=\left(0.2177\right)$

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

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

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

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

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