4. A grapefruit is launched out of a cannon at 17 m/s at 25° above horizontal. What is the time inthe air if the cannon was 5 m above the ground?

To find the time in the air, we can use the kinematic equations of motion.

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity of the grapefruit is given as 17 m/s at 25° above horizontal.

The horizontal component of the velocity (Vx) can be found using the formula: Vx = V * cos(θ), where V is the magnitude of the velocity and θ is the angle above horizontal.

Vx = 17 m/s * cos(25°) = 15.38 m/s (rounded to two decimal places)

The vertical component of the velocity (Vy) can be found using the formula: Vy = V * sin(θ).

Vy = 17 m/s * sin(25°) = 7.26 m/s (rounded to two decimal places)

Step 2: Determine the time taken for the grapefruit to reach its maximum height.

The time taken to reach the maximum height can be found using the formula: t = Vy / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

t = 7.26 m/s / 9.8 m/s^2 = 0.74 seconds (rounded to two decimal places)

Step 3: Determine the total time in the air.

Since the grapefruit reaches its maximum height halfway through its flight, the total time in the air is twice the time taken to reach the maximum height.

Total time in the air = 2 * 0.74 seconds = 1.48 seconds (rounded to two decimal places)

Therefore, the time in the air for the grapefruit is approximately 1.48 seconds.