You throw a water balloon hoping to hit your sibling. You throw it at an angle of 22.0° at 9.70 m/s. It ends up leaving your hand at a height of 0.700 m above the ground. It completely misses your sibling and breaks on the ground. How far did the water balloon travel?
Question
You throw a water balloon hoping to hit your sibling. You throw it at an angle of 22.0° at 9.70 m/s. It ends up leaving your hand at a height of 0.700 m above the ground. It completely misses your sibling and breaks on the ground. How far did the water balloon travel?
Solution
To solve this problem, we need to use the equations of motion. We can break down the initial velocity into its horizontal and vertical components, and then use these to find the time the balloon is in the air and the horizontal distance it travels.
Step 1: Find the initial horizontal and vertical velocities. The initial horizontal velocity (Vx) is given by Vcos(θ) = 9.70 m/s * cos(22.0°) = 8.96 m/s. The initial vertical velocity (Vy) is given by Vsin(θ) = 9.70 m/s * sin(22.0°) = 3.64 m/s.
Step 2: Find the time the balloon is in the air. We can use the equation of motion h = Vyt - 0.5gt^2, where h is the height above the ground, Vy is the initial vertical velocity, g is the acceleration due to gravity (9.81 m/s^2), and t is the time. We can rearrange this to find t = (Vy + sqrt(Vy^2 + 2gh))/g = (3.64 m/s + sqrt((3.64 m/s)^2 + 29.81 m/s^2*0.700 m))/9.81 m/s^2 = 0.53 s.
Step 3: Find the horizontal distance the balloon travels. The horizontal distance (d) is given by d = Vx*t = 8.96 m/s * 0.53 s = 4.75 m.
So, the water balloon traveled 4.75 m.
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