(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Projectile Motion question help

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