Answer to Question #96890 in Mechanics | Relativity for Musah fuseini

Question

A projectile is launched with speed Vo at an angle "theata" above the horizontal. The launch point is at a height h above the ground.
(a) show that if air resistance is neglected the horizontal range, R that the projectile travels before striking the ground is R = Vo*cos(theata)\g(VoSin"theat a"+√Vo^2Sin^2" theata"+2gh).
(b) what will be the expression for R if the launch point is at the ground level so that h = 0
(c)what

Answer 1

a) The maximum height the projectile will reach after it's launched from the height $h$:

$$H=h+\frac{(v_0\text{ sin}\theta)^2}{2g}=\frac{2gh+(v_0\text{ sin}\theta)^2}{2g}.$$

And this will take time

$$t_1=\frac{v_0\text{ sin}\theta}{g}.$$

Then it will fall back to the ground during

$$t_2=\sqrt{\frac{2H}{g}}=\frac{\sqrt{2gh+(v_0\text{ sin}\theta)^2}}{g}.$$

Meanwhile the projectile will move forward with constant velocity

$$v_x=v_0\text{ cos} \theta,$$

and of course it will cover the range

$$R=v_x(t_1+t_2)=\\=\frac{v_0\text{ cos} \theta}{g}(v_0\text{ sin}\theta+\sqrt{2gh+(v_0\text{ sin}\theta)^2}).$$

b) Substitute $h=0$ and get simple relation for the horizontal range:

$$R(0)=\frac{v_0^2\text{ sin} (2\theta)}{g}.$$

c) That is the answer to this problem.