A water balloon is launched vertically in the air from a height of 1 meter with an initial velocity of 30 meters per second. How high will the balloon be after 3 seconds? The acceleration due to gravity is -9.8 meters per second?. [?] meters

To find the height of the water balloon after 3 seconds, you can use the following equation of motion: \[ h = h_0 + v_0 t + \frac{1}{2} a t^2 \] Here, \( h_0 = 1 \) meter (initial height), \( v_0 = 30 \) meters/second (initial velocity), \( a = -9.8 \) meters/second² (acceleration due to gravity), and \( t = 3 \) seconds. Plugging in these values: \[ h = 1 + (30)(3) + \frac{1}{2}(-9.8)(3^2) \] Calculating it step-by-step: 1. \( (30)(3) = 90 \) 2. \( \frac{1}{2}(-9.8)(9) = -44.1 \) 3. \( h = 1 + 90 - 44.1 = 46.9 \) So, the height of the balloon after 3 seconds is approximately 46.9 meters.