Standing on the roof of a (42.0+a) m tall building, you throw a ball straight up with an initial speed of (14.5+b) m/s. if the ball misses the building on the way down, how long will it take from you threw the ball until it lands on the ground below? give your answer in seconds and round the answer to three significant figures.

First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:

2as = v² - u²

Because the final velocity v is 0 in such cases

s = -u²/2a; because both u and a are downwards, the negative sign cancels

s = 14.5² / 2*9.81
s = 10.72 meters

Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m

We will use the formula 
s = ut + 0.5at²

to find the time taken with the initial velocity u = 0.

55.72 = 0.5 * 9.81 * t²

t = 3.37 seconds

It should be d if not then c

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.