1.) Let's say that an an orange is launched upward at 22m/s. What is the displacement 3.5s later?

2.) When a volcano erupts, big rocks have been measured to reach heights of 50km (50,000m) What would the initial velocity need to be in order to reach such heights (neglect air resistance)?

To calculate the displacement of an object, you need to know the initial velocity (launch velocity in this case), the time interval, and any changes in acceleration.

For the first question:
1) Given that an orange is launched upward at 22m/s, we can assume the initial velocity (v0) is 22m/s.
2) The time interval (t) in this case is 3.5 seconds.
3) To calculate the displacement (Δy), you can use the following formula: Δy = v0 * t - (1/2) * g * t^2. Since the orange is launched upward, the acceleration due to gravity (g) is -9.8m/s^2 (negative because it acts in the opposite direction).
4) Substituting the given values into the formula, we get Δy = 22m/s * 3.5s - (1/2) * 9.8m/s^2 * (3.5s)^2.

For the second question:
1) The height reached by the rock is 50,000m.
2) Since the initial velocity (v0) is what we want to determine, we can denote it as v0.
3) We neglect air resistance in this case, so the only force acting on the rock is gravity. Therefore, the acceleration due to gravity is -9.8m/s^2.
4) The formula to calculate the maximum height achieved is: Δy = v0^2 / (2 * |g|), where |g| represents the absolute value of the acceleration due to gravity.
5) Substituting the given values into the formula, we get 50,000m = v0^2 / (2 * 9.8m/s^2).

To solve both equations, you can follow these steps:
1) Rearrange the equation to solve for the unknown variable.
2) Plug in the known values.
3) Calculate the values using a calculator or by hand.

Note: In both questions, make sure to pay attention to the direction of motion and the signs of the values to ensure accurate calculations.

a. h=vi*time-4.9 t^2

b. initial KE=final PE
1/2 m v^2=mgh
v^2=2*9.8*50,000
solve for v.