A ball of mass 2.0 kg is thrown vertically upward front the top of a building. The ball’s velocity v is given as a function of time t by the equation v(t)=R-St, where R=5 m/s and s=10 m/s^2. The positive direction is upward. What is the ball’s acceleration at time t= 1 s A. -10 m/s ^2
B. -5 m/s ^2
C. 0
D. 5 m/s^2
E. 10 m/s^2

Answer: 1.4 seconds which is the third optionexplanation: current is defined as the flow of charge per timetherefore, the equation that relates current, charge and time can be expressed as follows: i = δq / δtwhere: i is the current given = 1.8 ampereδq is the charge given = 2.5 coulombδt is the time required for the flow of the given charges to give the 1.8 ampere.substitute with the givens in the above equation to get the time as follows: 1.8 = 2.5 / δtδt = 2.5 / 1.8 = 1.38888 which is approximately 1.4 secondshope this was !