# A cannon barrel is elevated at an angle of 45°. It fires a ball with a speed of 250 m/s. (For the following questions, ignore air resistance.)

(a) What height does the ball reach?

m

(b) How long is the ball in the air?

s

(c) What is the horizontal range of the cannon?

km

## Vi = 250 sin 45 = 125 sqrt 2 initial up

u = 125 sqrt 2 forever horizontal

v = 125 sqrt 2 - 9.81 t

v at top = 0

9.81 t = 125 sqrt 2

solve for t at top

h = Vi t - 4.9 t^2

solve for h at top

total time in air = 2 t

d = 2 u t

## To answer these questions, we need to use the basic principles of projectile motion.

Let's consider the following information:

- The initial speed of the ball is 250 m/s.

- The cannon barrel is elevated at an angle of 45°.

(a) What height does the ball reach?

To find the maximum height reached by the ball, we need to use the formula for vertical displacement:

h = (v^2 * sin^2θ) / (2g)

Where:

- h is the maximum height reached by the ball

- v is the initial speed of the ball (250 m/s)

- θ is the angle of elevation (45°)

- g is the acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the values into the formula:

h = (250^2 * sin^2(45°))/(2 * 9.8)

Simplifying:

h = (62500 * 0.5) / 19.6

h ≈ 1596 m

Therefore, the ball reaches a height of approximately 1596 meters.

(b) How long is the ball in the air?

To find the total time the ball is in the air, we need to consider the time it takes for the ball to reach its maximum height and then fall back down. Since the launch angle is 45°, the time taken to reach the maximum height is the same as the time taken for the ball to return to the same height.

The formula to calculate the time of flight is given by:

t = (2v * sinθ) / g

Where:

- t is the time of flight

- v is the initial speed of the ball (250 m/s)

- θ is the angle of elevation (45°)

- g is the acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the values into the formula:

t = (2 * 250 * sin(45°)) / 9.8

Simplifying:

t = (500 * 0.7071) / 9.8

t ≈ 35.99 s

Therefore, the ball is in the air for approximately 36 seconds.

(c) What is the horizontal range of the cannon?

To find the horizontal range, we need to calculate the horizontal distance traveled by the ball before hitting the ground. The horizontal range is given by:

R = (v^2 * sin2θ) / g

Where:

- R is the horizontal range

- v is the initial speed of the ball (250 m/s)

- θ is the angle of elevation (45°)

- g is the acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the values into the formula:

R = (250^2 * sin(2*45°))/(9.8)

Simplifying:

R = (62500 * 1) / 9.8

R ≈ 6377 m

Therefore, the horizontal range of the cannon is approximately 6377 meters or 6.377 kilometers.