distance a cannonball travels when shot horizontally with a velocity of 28.3 at 30 degrees? I'm not sure what formula I need to use. Please help.
To find the distance a cannonball travels when shot horizontally with a velocity of 28.3 m/s at an angle of 30 degrees, you can use the following steps:
1. Split the initial velocity into horizontal and vertical components. The horizontal component (Vx) represents the velocity in the x-axis (horizontal direction) and remains constant throughout the projectile's motion. The vertical component (Vy) represents the velocity in the y-axis (vertical direction) and changes due to the acceleration of gravity.
Vx = V * cos(theta)
Vy = V * sin(theta)
Where:
V is the initial velocity (28.3 m/s)
theta is the launch angle (30 degrees)
cos(theta) is the cosine of the launch angle
sin(theta) is the sine of the launch angle
2. Find the time taken for the cannonball to reach its highest point. At the highest point, the velocity in the vertical direction becomes zero. You can use the equation:
t = Vy / g
Where:
t is the time taken
g is the acceleration due to gravity (approximately 9.8 m/s^2)
3. Calculate the total time of flight by doubling the time taken to reach the highest point:
total time of flight (T) = 2 * t
4. Finally, find the horizontal distance traveled by the cannonball using the equation:
distance = Vx * T
Substituting the values calculated earlier, you can compute the distance traveled by the cannonball when shot horizontally with a velocity of 28.3 m/s at an angle of 30 degrees.