A small steel ball bearing with a mass of 28.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.25 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 34.0 deg from the horizontal.

Answer 1

energy in spring: m*g*h
velocity of the ball= sqrt 2gh check that.we will use this later.

now calculate the time in air launched at 34deg

hf=hi+Vi*sin34*time-4.9time^2
solve for time in air, t.

horizontal distance= Vi*cos34*t

Answer 2

my answer for this question came out to be 1.16 m. However when I plug it into the Physics homework program it say incorrect.

My equation for t was:
1.25 m = 0m + 4.952sin34t - 4.9t^2 and then I did the quadratic equation and got t= .28258

I then did the horizontal distance equation using 4.952 as my Vi.

What am I doing wrong where I still don't get the correct answer?

Answer 3

The way to solve this is to use potential and kinetic energy and the formula R = (v^2/g)sin2(theta).
mgh=(1/2)mv^2
v=sqrt(2gh)
R=(sqrt(2gh)^2/9.8)sin2(theta)
= ((4.95)^2/9.8)sin(2*34)
=3.27m