An algorithm has been written in pseudocode to input 50 numbers and output the sum. a. Find the error(s) in the pseudocode and suggest a correction.
b. Change the corrected pseudocode that you suggested in the solution above such that it doesn’t allow the user to enter any negative numbers. If a negative number is entered it should display an error message as well as ask for input again.
c. Dry run the solution you suggested in part b above and create a trace table for it. You may choose test data of your choice. (It should contain both positive and negative numbers)
d. Convert the solution you suggested in part b to its equivalent flowchart.

a) h = 1/2 gt² b) 99 m/s c) 14.1 m/s

Explanation:

a) As the only force acting on the canister once released is gravity, as it is constant, we can use one of the kinematic equations, as follows:

h = h₀ + v₀t + 1/2a t²

As we are told that the canister is simply released, it means that v₀ =0.

If we choose the direction of the acceleration (downward) as positive ,and we select the height at which it was released as our origin, so h₀ =0, the final expression for height is as follows:

h = 1/2 g t²

b) Combining this equation with the expression of t, from the definition of acceleration as the rate of change of velocity, we arrive to this expression:

vf² - v₀² = 2 g h  

As v₀ = 0, we have v₀ = √2.g.h = √2.9.8.500 m²/s² = 99 m/s

c) In order to be able to break the canister, impact speed must be larger than 100 m/s.

So, we can use the same equation as above, putting vf=100 m/s, and solving for v₀, as follows:

v₀² = vf² - 2.g.h = 100² m²/s² - 2.9.8.500 m²/s² = 200 m²/s²

v₀ = √200 m²/s² = 14.1 m/s

For any value of v₀ just barely larger than this, the canister will be broken.