I need help with these. I'm so confused! Thanks!

1.) An F-14 Tomcat fighter jet goes from rest to a speed of 75m/s in 2.5 s. What is the acceleration?

2.) Preston tossed a red ball upward and it reaches a maximum height of 3.0m What is the final velocity when it returns to Preston's hand?

acceleration = change in velocity / change in time

same as speed on the way up, but down not up or in other words it falls from 3 meters up

m g h = (1/2) m v^2
so
v = -sqrt(2 g h) = -sqrt ( 2 * 9.81 * 3)

Sure! I can help you with those.

1) To find the acceleration, we can use the formula: acceleration (a) = change in velocity (Δv) / time (t). In this case, the change in velocity is from rest (0 m/s) to 75 m/s, and the time is 2.5 seconds.

So, the formula becomes: a = (75 m/s - 0 m/s) / 2.5 s

Simplifying the equation, we get: a = 75 m/s / 2.5 s

By performing the division, we find that the acceleration is 30 m/s².

2) When the ball returns to Preston's hand, its final velocity will be the same as its initial velocity (since the ball will have the same magnitude of velocity but in the opposite direction). In this case, the initial velocity is the velocity at the maximum height (when the ball stopped and started moving downward) and the magnitude was 0 m/s.

Therefore, the final velocity when the ball returns to Preston's hand is 0 m/s.