A bullet is fired at 45 degrees with respect to horizontal with a velocity of 50 m/s. How long is the bullet in the air? What is the range?

To find the time the bullet is in the air, we can use the formula for the time of flight of a projectile. The formula is:

t = (2 * v * sinθ) / g

where:
t = time of flight
v = initial velocity of the bullet
θ = angle of projection (in this case, 45 degrees)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the values we have:
v = 50 m/s
θ = 45 degrees
g = 9.8 m/s^2

t = (2 * 50 * sin45°) / 9.8

To simplify the calculation, we can use the fact that sin 45 degrees is equal to √2 / 2:

t = (2 * 50 * (√2 / 2)) / 9.8

We can further simplify the equation:

t = (50 * √2) / (9.8)

t ≈ 3.18 seconds (rounded to two decimal places)

So, the bullet is in the air for approximately 3.18 seconds.

To find the range, we can use the formula:

R = (v^2 * sin(2θ)) / g

where:
R = range (horizontal distance covered by the bullet)
v = initial velocity of the bullet
θ = angle of projection (in this case, 45 degrees)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the values we have:
v = 50 m/s
θ = 45 degrees
g = 9.8 m/s^2

R = (50^2 * sin(2 * 45°)) / 9.8

We can simplify the equation using the double angle identity sin(2θ) = 2sinθcosθ, and since sin 45 degrees and cos 45 degrees are equal to √2 / 2:

R = (2500 * (2 * (√2 / 2) * (√2 / 2))) / 9.8

We can further simplify the equation:

R = (2500 * (√2 * √2)) / (9.8)

R = (2500 * 2) / (9.8)

R ≈ 510.20 meters (rounded to two decimal places)

So, the range of the bullet is approximately 510.20 meters.