- #1

vaironl

- 34

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The following problem wasn't assigned to me, it was used for last semesters physics class at my college. However, I feel it is interesting enough and I can't seem to find the correct answer.

A ball is launched from the top of a 28-m high vertical cliff at an angle of 31° . Ignoring the effects of air resistance, if the ball is to hit a target on the ground a horizontal distance 65-m away from the edge of the cliff, with what initial speed must it be launched?

This is the problem and I haven't excluded or included information.

## Homework Equations

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[tex]V_x= V_{x0} + a_x*t[/tex]

[tex]x = x_0 + v_{x0}*t + (1/2)a_x*t^2[/tex]

[tex]x - x_0 = t * (v_{x0} + v_x)/(2)[/tex]

[tex]a_y = m*g[/tex]

## The Attempt at a Solution

[/B]

I first attempted to split the initial velocity to its components

[tex]V_{x0} = cos(31) * V[/tex], [tex]Vy_{y0} = sin(31)*V[/tex]

I assumed there's no acceleration in the horizontal direction and noticed that Vx = V_(x0)

Now I feel like I don't have enough given information to get the initial velocity since all kinematic equations given depend on initial velocity components which is not given. Neither are we given final velocity components.

At some point I noticed [tex]x - x_0 = t * \frac{v_{x0}+ vx}{2}[/tex] can be manipulated in the following way

[tex]v_x = v_{x0}[/tex]

[tex]x-0 =\frac{v_{x0}+v_{x0}}{2}t [/tex]

[tex]x=\frac{2v_{x0}}{2}t [/tex]

[tex]v_{x0} = \frac{x}{t} [/tex]

x = 65, but t seems more difficult to find.

Am I even heading somewhere here?