Question

A water balloon slingshot launches its projectiles essentially from ground level at a speed of 24.0m/sm/s . (You can ignore air resistance.) At what angle should the slingshot be aimed to achieve its maximum range? If shot at the angle you calculated in Part A, how far will a water balloon travel horizontally? For how long will the balloon be in the air?

Answer 1

To achieve maximum range, the water balloon slingshot should be aimed at a 45° angle, which will allow the projectile to travel approximately 58.8 meters horizontally and be in the air for around 3.49 seconds.

Step-by-step explanation:

To achieve maximum range for a projectile launched from ground level without air resistance, it should be launched at an angle of 45° above the horizontal. This angle ensures that the horizontal and vertical components of the initial velocity are equal, optimizing the range according to the principles of projectile motion.

To determine how far a water balloon will travel horizontally when launched at this angle, we use the following equations of projectile motion:

• Range (R) = (v² × sin(2θ)) / g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (9.8 m/s²).
• Time of flight (T) = (2 × v × sin(θ)) / g.

The water balloon will travel approximately 58.8 meters horizontally and will be in the air for roughly 3.49 seconds.