A clown is shot from a cannon with a speed of 25 m/s at an angle of 35 degrees. If his landing is at the same level as his take-off point, then how long is he in the air?
To determine how long the clown is in the air, we first need to break down the initial velocity into horizontal and vertical components. The horizontal component represents the clown's velocity in the x-direction (parallel to the ground), while the vertical component represents the velocity in the y-direction (perpendicular to the ground).
Given the initial speed of 25 m/s and the launch angle of 35 degrees, we can calculate the horizontal and vertical components using trigonometric functions.
Horizontal component:
Vx = V * cos(theta)
Vx = 25 m/s * cos(35 degrees) = 20.4 m/s
Vertical component:
Vy = V * sin(theta)
Vy = 25 m/s * sin(35 degrees) = 14.3 m/s
Since the clown lands at the same level as his take-off point, we can assume there is no change in the vertical displacement. The time of flight (t) can be determined by using the equation:
t = 2 * (Vertical component) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
t = 2 * 14.3 m/s / 9.8 m/s^2
t ≈ 2.92 seconds
Therefore, the clown is in the air for approximately 2.92 seconds.