A bullet is fired from a gun at 300 m/s. It hits the ground at 3s later. At what angle (in degrees) above the horizon was the bullet fired? (Assume no air resistance)

Well, I guess the bullet really hit rock bottom! But let's get serious for a second. We can use the equation of motion to find the angle.

The vertical velocity of the bullet starts from zero, experiences free-fall acceleration, and then comes to a stop at the ground. So, we can use the equation:

h = (1/2)gt^2

Where h is the vertical displacement, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.

Substituting the values, we get

h = (1/2)(9.8)(3)^2 = 44.1 m

Now, let's use another equation of motion to find the initial vertical velocity:

v = u + gt

Since u (initial vertical velocity) is 0, we have:

300*sinθ = 9.8*3

Simplifying this equation, we find:

θ = sin^(-1)((9.8*3)/300)

Calculating this in our calculators, we get:

θ ≈ 0.566 rad

But we need the angle in degrees, not radians. Converting from radians to degrees, we have:

θ ≈ 32.46 degrees (approximately)

So, drumroll, please... the bullet was fired at an angle of approximately 32.46 degrees above the horizon!

To determine the angle at which the bullet was fired, we can use the kinematic equation for vertical motion:

h = v0y * t - (1/2) * g * t^2

where:
h = vertical displacement (in this case, the height of the ground, which is 0)
v0y = initial vertical velocity (which we need to find)
t = time of flight (3 seconds)
g = acceleration due to gravity (9.8 m/s^2)

Since the bullet is fired horizontally, its initial vertical velocity is 0. Therefore, the equation becomes:

0 = 0 * 3 - (1/2) * 9.8 * (3^2)

Simplifying the equation:

0 = -14.7 * 9

Since the vertical displacement is 0, the bullet hits the ground when the equation equals 0. Solving for g:

0 = -14.7 * t^2

t^2 = 0

Since the square of time cannot be negative, the only valid solution is t = 0. Therefore, the bullet was never fired.

Please double-check your values and make sure they are accurate.

To find the angle at which the bullet was fired, we can use the equations of motion to determine the vertical and horizontal components of the bullet's velocity.

Let's consider the vertical motion first. We know that the acceleration due to gravity is acting downward, and the final displacement is 0 since the bullet hits the ground. Therefore, we can use the following equation:

s = ut + 0.5at^2

where s is the vertical displacement, u is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.

Since the bullet was fired vertically, the initial vertical velocity (u) is equal to the initial velocity of the bullet. Therefore, we can write:

0 = 300*t + 0.5*(-9.8)*t^2

Simplifying this equation, we get:

-4.9t^2 + 300t = 0

Now, solve this quadratic equation to find the time (t) when the bullet hits the ground. We can factor out t:

t(-4.9t + 300) = 0

Either t = 0 (initial time) or -4.9t + 300 = 0. Since t cannot be zero in this case, we solve the other equation:

-4.9t + 300 = 0
-4.9t = -300
t = -300 / -4.9
t ≈ 61.2 seconds

Since the bullet hits the ground after 3 seconds, the negative value obtained above is not relevant. Therefore, the time taken to hit the ground is approximately 3 seconds.

Now, let's determine the horizontal distance covered by the bullet. We can use the equation:

s = ut

where s is the horizontal displacement, u is the initial horizontal velocity, and t is the time.

The horizontal velocity remains constant throughout the motion because there is no horizontal acceleration (we assume no air resistance). Therefore, the horizontal displacement is:

s = u*t

Since we know the initial velocity (u = 300 m/s) and the time (t = 3 seconds), we can calculate the horizontal displacement as:

s = 300 * 3
s = 900 meters

Now, we can use trigonometry to find the angle above the horizon at which the bullet was fired. The horizontal displacement (900 meters) and the initial vertical velocity (300 m/s) form a right-angled triangle.

tan(θ) = vertical displacement / horizontal displacement
tan(θ) = 300 / 900

Using a scientific calculator or trigonometric tables, we can determine the value of θ:

θ ≈ 18.4°

Therefore, the bullet was fired at an angle of approximately 18.4 degrees above the horizon.

The time of the projectile motion is

t=2v₀sinα/g

sinα=gt/2v₀ =9.8•3/2•300=0.05
α=arcsin0.05=2.8°