# Projectile Motion question help

1. Sep 16, 2008

### crazy_nuttie

1. The problem statement, all variables and given/known data

A helicopter is rising vetically at a uniform velocity of 14.7 m/s. When it is 196 m from the ground, a ball is projectted from it with a horizontal velocity of 8.5 m/s with respect to the helicopter. Calculate:
a) When the ball will reach the ground
b) Where it will hit the ground
c) What its velocity will be when it hits the ground

Answers are: 8.0 s, 68 m, 64 m/s 82 degrees below horizontal

2. Relevant equations

c² = root of a² + b²

d = v1 t + 1/2 a (t)²

3. The attempt at a solution

So i drew out the path of the projected ball and used Pythagorean theorem to solve the 2 vectors of 14.7 [N] and 8.5 [E]

and i get 16.9 m/s [N 60 E]

I used d = v1 t + 1/2 a (t)²
to get distance

However, my answer comes out incorrect (by a lot)

so i used:

V2 ^ 2 - V1 ^2 = 2 a d

285 / 2(-9.8m/s^2) = d

d = 15 m

However, That is not the correct answer.

Looking for someone to point me in the right direction, thanks.

2. Sep 16, 2008

### Feldoh

For part one, you must realize that the only force (neglecting air-resistance) is gravity, which only occurs in the y (downward) direction. So the time to reach the ground is independent of the horizontal velocity!

Another thing to remember that when dropped the ball has an initial velocity that is not 0. The helicopter that dropped the ball is moving upward!

Does that give you a nudge in the right direction?

3. Sep 16, 2008

### crazy_nuttie

I dont quite understand what you mean. I understand that gravity is the only force acting upon Vy, but how do you get the time as to when it will reach the ground?

4. Sep 16, 2008

### Feldoh

You all ready have the equation:

d = d0 + v0t - 1/2at^2