Question

A person is launching an object out of a cannon at 59 degrees, the object lands at the same height, 497m away.
Find initial speed. (assume standard coor. system and no wind or air resistance)

Answer 1

Tough question.

\( \color{green}{\text{horizontally: } t = \frac{497}{v_i \cos 59^{\circ}} \\ \text{vertically, to reach max height at } \dfrac{1}{2}t: 0 = v_i \sin 59^{\circ} + \dfrac{1}{2} (9.8)t \\ \text{combining for t: } \frac{v_i \sin 59^{\circ}}{(1/2)(9.8)} = \frac{497}{v_i \cos 59^{\circ}} }\)

Solve for the initial velocity.

Answer 2

could you revisit this with me. every time i work it out i get a number around 5,200 and that can not be right at all.

Answer 3

Let the initial vel. be v. Verticlal vel = v sin 59 and horizontal vel = v cos 59
Vertical displacement = 0
so, 0 = v sin 59 t - g t^2 /2
t = 0 and t = 2 v sin 59/g
horizontal displacement = v cos 59 * t
497 = v cos 59 * 2v sin 59/g
497*g/0.86 = v^2
Now calculate v.