(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A ball is launched from a cliff 1 meter high. At the bottom of the cliff, the ground is not flat, it slants downward at a 45 degree angle as show. The ball's initial; velocity is 10 m/s and makes a 30 degree angle with the horizontal. Find x, the horizontal displacement of the ball from the edge of the cliff when it lands.

Ans: 24.85 m

Please see the picture

[PLAIN]http://img36.imageshack.us/img36/8817/bookscanstation20110918.jpg [Broken]

2. Relevant equations

v^2 = v_0^2 + 2 a delta x

v = v_0 + a t

x = x_0 + v_0 t + 1/2 a t^2

3. The attempt at a solution

I set point were the ball is originally to be represented by the Cartesian coordinates (0,0) and t = 0. I choose to the right of this point to be positive, above this point to be positive, and below this point to be negative.

I solved for the maximum height of the ball

v^2 = v_0^2 + 2a delta y

delta y = (v^2 - v_0^2)/(2a)

Used the fact that the velocity at the top of the ball is zero, the initial velocity in the y direction is 10 sin(30) = 5 m/s, and the acceleration due to gravity is about 9.8 m/s^2 downwards

delta y = - 5^2/(2*-9.8) = 1.276 meters

I solved for the time it takes for the ball to get to this point

v = v_0 + at

t = (v - v_0)/a

t = -5/-9.8 = .510 seconds

I than solved for the horizontal displacement at this point

If we ignore any type of acceleration in the horizontal direction

x = v_0 t

v_0 in the x direction is 10 cos(30) = 5sqrt(3)

x = 5 sqrt(3) t

x = 5 sqrt(3) .510 = 4.417 meters

I than solved for how far below the ground is directly below this point x = 4.417 level to the point that can be represented to describe the point right before the ground starts slanting downwards (0,-1) using trigonometric functions

4.147 tan(45) = 4.147 meters

I than concluded that the equation that can represent the vertical fall at any point from y=-1 to be

y = 5sqrt(3)t

Using the general equation for displacement for the vertical direction

y = y_0 + v_0 t + 1/2 a t^2

and setting y (the point that represents the final height of the ball) to be the opposite of 4.417 + 1 + 5 sqrt(3) t

or simple

-(5.417 + 5sqrt(3) t)

I than set this equal to the right hand side of the equation (the initial velocity in the vertical direction of the ball at the top of it's trajectory is zero) and using the point 1.276 (the point that can be used to describe the height of the ball at the top of it's trajectory) I got

-(5.417 + 5sqrt(3) t) = 1.276 - 4.9 t^2

I than used the solve feature on my calculator and got that

t = 2.349 seconds

I than plugged this into the equation that represents the horizontal displacement of the ball

x = 5 sqrt(3) 2.349

x = 20.343 meters

The answer is 24.85 meters and I don't see what I'm doing wrong. Thanks for any help that you can provide.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Motion - Projectile

**Physics Forums | Science Articles, Homework Help, Discussion**