A 0.0309-kilogram ball is stuffed into a spring gun and shot up into the air at some angle. The gun's spring has a force constant of 92.5 N/m and is initially compressed 0.138 m. When the ball reaches maximum height, it has a horizontal velocity with a speed of 2.23 m/s. How high above its initial location in the gun does the ball soar?
To find the height above its initial location that the ball soars, we need to use the principles of projectile motion. Here's how to solve the problem step by step:
Step 1: Determine the initial velocity of the ball.
We know that the ball is shot vertically out of the spring gun. The initial vertical velocity (vy) can be determined using the conservation of mechanical energy. At maximum height, all the potential energy of the compressed spring is converted into gravitational potential energy. Therefore, we can equate the initial potential energy to the final gravitational potential energy.
The initial potential energy can be calculated as the potential energy stored in the compressed spring:
Ep = (1/2) k x²
Where k is the force constant of the spring (92.5 N/m) and x is the compression of the spring (-0.138 m as it is compressed).
Ep = (1/2) * 92.5 N/m * (-0.138 m)²
Ep = 0.902 Joules
At maximum height, all the potential energy is converted into kinetic energy. Therefore, the final kinetic energy is equal to the initial potential energy:
Ek = (1/2) m vy²
0.902 Joules = (1/2) * 0.0309 kg * vy²
vy² = 58.50 m²/s²
Taking the square root of both sides:
vy ≈ √58.50 m²/s²
vy ≈ 7.64 m/s
Step 2: Determine the time taken to reach maximum height.
At maximum height, the vertical component of the velocity (vy) is 0 since the object momentarily stops moving upward before falling back down. We can use this information to determine the time taken to reach maximum height.
vy = gt
Where g is the acceleration due to gravity (-9.8 m/s²) and t is the time taken to reach maximum height.
0 = -9.8 m/s² * t
t ≈ 0.78 s
Step 3: Calculate the height reached by the ball.
The height reached by the ball can be calculated using the formula:
h = vy * t + (1/2) * g * t²
h = (7.64 m/s) * (0.78 s) + (1/2) * (-9.8 m/s²) * (0.78 s)²
h ≈ 5.99 m
Therefore, the ball soars approximately 5.99 meters above its initial location in the gun.