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.

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

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

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?

To calculate the horizontal distance the ball would travel, we need to consider the projectile motion of the ball when it is released from the spring.

First, let's find the initial velocity of the ball when it is released vertically. To do this, we will use the principle of conservation of mechanical energy.

The initial potential energy of the ball when it is at the maximum height can be calculated as:
Potential energy = mass * gravity * height
Potential energy = 0.028 kg * 9.8 m/s^2 * 1.25 m = 0.3435 J

Since the spring is compressed, this potential energy is converted to elastic potential energy in the spring when released. Thus, the elastic potential energy can be calculated as:
Elastic potential energy = 0.3435 J

Next, we can find the initial velocity of the ball using the equation for elastic potential energy:
Elastic potential energy = 0.5 * mass * velocity^2
0.3435 J = 0.5 * 0.028 kg * velocity^2

Solving for velocity:
velocity^2 = (0.3435 J * 2) / 0.028 kg
velocity^2 = 24.5357 m^2/s^2
velocity ≈ 4.95 m/s

Now that we have the initial vertical velocity, we can find the horizontal distance the ball would travel when the spring is aimed at an angle of 34.0 degrees from the horizontal.

The horizontal distance can be calculated using the equation for projectile motion:
horizontal distance = (initial horizontal velocity * time of flight)

Since there is no horizontal acceleration, the time of flight will be the same for both the vertical and angled release.

We can calculate the time of flight using the equation for vertical motion:
time of flight = 2 * (vertical initial velocity) / g
time of flight = 2 * 4.95 m/s / 9.8 m/s^2 ≈ 1.01 s

Now, let's find the horizontal velocity component when the ball is released at an angle of 34.0 degrees.

horizontal initial velocity = initial velocity * cos(angle)
horizontal initial velocity = 4.95 m/s * cos(34°) ≈ 4.10 m/s

Finally, we can calculate the horizontal distance using the equation:
horizontal distance = horizontal initial velocity * time of flight
horizontal distance ≈ 4.10 m/s * 1.01 s ≈ 4.14 m

Therefore, the ball would travel approximately 4.14 meters horizontally when the spring is aimed 34.0 degrees from the horizontal.