kid shoots a BB gun directly upward at a ock of birds. The initial velocity of the BB is21 m/s. If the BB hits a bird at less than 1 m/s then it will not harm the bird. Calculate the minimumheight above the gun at which the birds may safely y. Use a magnitude of acceleration of 9.8m

22.5 m

Explanation:

Using v² = u² - 2gy where u = initial velocity of BB = 21 m/s, v = final velocity of BB = 1 m/s (since this is the required speed of BB in which it will not harm the birds), g = acceleration due to gravity = 9.8 m/s² and y = minimum height of BB above the gun at which the birds may safely fly.

Substituting the values of the variables into the equation, we have

v² = u² - 2gy

(1 m/s)² = (21 m/s)² - 2(9.8 m/s²)y

collecting like terms, we have

(1 m/s)² - (21 m/s)² = - 2(9.8 m/s²)y

1 m²/s² - 441 m²/s² = -(19.6 m/s²)y

simplifying, we have

- 440 m²/s² = -(19.6 m/s²)y

dividing through by -19.6 m/s², we have

y = - 440 m²/s² ÷ -19.6 m/s²

y = 22.45 m

y ≅ 22.5 m

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

im srry but ur wrong the answer is d a smooth rock, some red dirt, and an empty bird's nest.