Answer:
The balloon needs a vertical velocity of approximately 22.14 m/s to reach the window
Explanation:
The given parameters are;
The destination of the designed water balloon we shoot = The 3rd story window
The height of the 3rd story window which the water balloon should reach = 25 m
Therefore, we have, the equation of motion of the water balloon given as follows;
v² = u² - 2 × g × s
Where;
[tex]u_y[/tex] = The initial vertical velocity of the balloon
[tex]v_y[/tex] = The final vertical velocity of the balloon = 0 m/s
g = The acceleration due to gravity = 9.8 m/s
s = The height the balloon is intended reach at the final velocity becomes 0 m/s = The height of the 3rd story window
∴ s = 25 m
Substituting the values, we have;
0² = [tex]u_y[/tex]² - 2 × 9.8 × 25
[tex]u_y[/tex]² = 2 × 9.8 × 25 = 490
[tex]u_y[/tex] = √490 = 7·√10
The initial vertical velocity of the balloon = [tex]u_y[/tex] = 7·√10 m/s
Therefore, the balloon needs a vertical velocity of 7·√10 m/s which is approximately 22.14 m/s to reach the window.
