Answer to Question #95865 in Mechanics | Relativity for Needed a really ASAP help.

Question

What initial velocity would be necessary for a projectile to be launched off a 50m high cliff so that it would produce a range of 473.67m and have a total time aloft of 8.04s. (g= 10N/kg)

Answer 1

Horizontal component of velocity:

$$v_x=\frac{R}{t}=\frac{ 473.67}{8.04}=58.91\frac{m}{s}$$ $$0=h+v_yt-0.5gt^2$$

$$0=50+v_y(8.04)-0.5(10)(8.04)^2$$

Vertical component of velocity:

$$v_y=33.98\frac{m}{s}$$

The magnitude:

$$v=\sqrt{58.91^2+33.98^2}=68.0\frac{m}{s}$$

The direction:

$$\theta=\arctan{\frac{33.98}{58.91}}=30.0\degree$$