A toy gun uses a spring to shoot a ball with a mass of 5.0 g. The spring is “loaded” by compressing it 5.0 cm and

has a spring constant of 288 N/m. The end of the spring is at the end of the gun (the ball does not travel any
distance inside the gun after the spring) and the toy gun is held at a height of 1.5 m above the ground. If the gun is aimed horizontally, how far can it shoot the ball? If it is aimed up 45 degrees how will the initial velocity change? How will the initial velocity change if shot straight upward, and how high would it go?

I really appreciate any assistance you can give me for this, thanks!

a. find out the spring energy before shooting.

b. find out the initial ball velocity from its KEnergy.
c. find out how long it takes the ball to fall 1.5 meters,
distance=velocity*time

c1!@45: find out initila horizontal velocity, and initial vertical velocity.
Intialvertical=vi*sin45
initialhorizonal=vi*cos45

time in air:
hf=hi+vi'*t -4.9 t^2
hf=-1.5, hi=0
solve for t (use quadratic formula)
then horizontal distance=vi*cos45*timeinair.

To find the distance the ball can be shot when the gun is aimed horizontally, we can use the concept of potential energy and mechanical energy.

First, let's calculate the potential energy of the compressed spring. The potential energy stored in a spring is given by the equation:

PE = (1/2)kx^2

where PE is the potential energy, k is the spring constant, and x is the compression of the spring. In this case, the spring constant is 288 N/m and the compression is 0.05 m (converted from 5.0 cm). Plugging these values into the equation, we get:

PE = (1/2)(288 N/m)(0.05 m)^2
= 0.36 J

Now, let's consider the conversion of potential energy into kinetic energy. When the spring is released, the potential energy is converted into kinetic energy. The kinetic energy of an object can be calculated using the equation:

KE = (1/2)mv^2

where KE is the kinetic energy, m is the mass of the ball, and v is its velocity. In this case, the mass of the ball is 0.005 kg (converted from 5.0 g).

Since the ball starts from rest (initial velocity, v = 0), the initial kinetic energy is also 0. Therefore, the entire potential energy of the spring is converted into kinetic energy:

PE = KE

0.36 J = (1/2)(0.005 kg)v^2

Simplifying the equation:

v^2 = (2 * 0.36 J) / 0.005 kg
v^2 = 72 m^2/s^2

Taking the square root of both sides:

v = √72 m/s
v ≈ 8.49 m/s

Now we have the initial velocity of the ball when the gun is aimed horizontally. To calculate the distance the ball can travel, we can use the equation for horizontal projectile motion:

d = v * t

where d is the distance, v is the initial velocity, and t is the time of flight. Since the ball is shot horizontally and there is no vertical acceleration, the time of flight can be determined using the equation:

t = 2h / g

where h is the height above the ground and g is the acceleration due to gravity (approximately 9.8 m/s^2).

In this case, the height above the ground is 1.5 m. Plugging these values into the equation, we get:

t = (2 * 1.5 m) / 9.8 m/s^2
t ≈ 0.306 s

Now we can calculate the distance:

d = (8.49 m/s) * (0.306 s)
d ≈ 2.60 m

Therefore, when the gun is aimed horizontally, the ball can travel approximately 2.60 meters.

Now let's consider the scenario when the gun is aimed up at an angle of 45 degrees. The initial velocity will change because the velocity has two components: horizontal and vertical.

The horizontal component of the initial velocity remains the same as before: 8.49 m/s.

The vertical component of the initial velocity can be determined using the equation:

v_y = v * sin(θ)

where v_y is the vertical component of the initial velocity and θ is the angle of elevation (45 degrees in this case).

Plugging in the values, we get:

v_y = (8.49 m/s) * sin(45°)
v_y ≈ 6.00 m/s

Therefore, the initial velocity of the ball when the gun is aimed up at an angle of 45 degrees is approximately 8.49 m/s horizontally and 6.00 m/s vertically.

Finally, let's consider the scenario when the gun is shot straight upwards. In this case, the initial velocity only has a vertical component, and the horizontal component is 0.

The initial velocity can be determined using the equation:

v_y = v * cos(θ)

where v_y is the vertical component of the initial velocity and θ is the angle of elevation (90 degrees in this case).

Plugging in the values, we get:

v_y = (8.49 m/s) * cos(90°)
v_y ≈ 0 m/s

Therefore, the initial velocity of the ball when the gun is shot straight upwards is approximately 0 m/s horizontally and 8.49 m/s vertically.

To calculate the maximum height the ball will reach when shot straight upwards, we can use the equation for vertical projectile motion:

h_max = (v_y^2) / (2 * g)

Plugging in the values, we get:

h_max = (8.49 m/s)^2 / (2 * 9.8 m/s^2)
h_max ≈ 3.65 m

Therefore, when shot straight upwards, the ball will reach a maximum height of approximately 3.65 meters.