Atennis ball on mars, where the acceleration due to gravity is 0.379g and air resistance is negligible, is hit directly upward and returns to the same level 8.5 s later. (a) how high above its original point did the ball go? (b) how fast was it moving just after being hit? (c) sketch clear graphs of the ballâs vertical position, vertical velocity, and vertical acceleration as functions of time while itâs in the martian air

 To solve this, we use the formula:

y = v0 t + 0.5 a t^2

where y is distance, v0 is initial velocity, t is time and a is acceleration

Since we know that total time is 8.5 seconds, hence going up must be 4.25 s and going down is 4.25 s.

a = 0.379 g = 0.379 (9.8 m/s^2) = 3.7142 m/s^2

going up:

y = v0 (4.25) - 0.5 (3.7142) (4.25)^2

y = 4.25 v0 – 33.5439                       --> eqtn 1

going down:

y = 0 (4.25) + 0.5 (3.7142) (4.25)^2

y = 33.5439

y = 33.5439 m

Calculating for v0 from equation 1:

33.5439 = 4.25 v0 – 33.5439

4.25 v0 = 67.0877

v0 = 15.78535 m/s

answers:

a. y = 33.5439 m

b. v0 = 15.78535 m/s

c.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j

1 is a answer of a question