How would I do this problem?

Question

Carol throws a softball upward from a height of 3 ft above the ground and with an initial velocity of 24 ft/second.

a. What is the maximum height the ball reaches?
b. When does the ball reach its maximum height?
c. When does the ball hit the ground?

keep in mind I want to know HOW to do this not just the answer.

Answer 1

Use the following formulas:

Vfinal = Vinitial + at and d = Vinitail*t+0.5*a*t^2

0 = 24 - 9.8t
t = 2.45 s, this is the time it takes to reach max height (b)

d = Vinitial*t - 0.5 a*t^2
d = 24 * 2.45 - 0.5 * 9.8 * 2.45 ^2
d = 29.4, so max height = 29.4 + 3 = 32.4 (a)

d = Vinital*t + 0.5*a*t^2
32.4 = 0*t + 0.5 * 9.8*t^2
t = 2.57 (c)

Answer 2

a=dv/dt = -32 <-- acceleration due to gravity in ft/sec^2
v = -32t +C
When t = 0, v = 24, so C =24, so
v = -32t+24
s = dv/dt
s = -16t^2 +24t + C
When t = 0, s = 3, so
s = -16t^2 +24t +3
Ball reaches max height when t = -b/2a = -24/(2*-16) = 3/4sec
Max height = -16(3/4)^2 +24(3/4) +3 = 12 feet
t = [-24 +/- sqrt(24^2-4(-16)(3))]/(2(-16)
t = [-24 +/- 16sqrt(3)]/-32
t =( 24+16sqrt(3))/32 = approx 1.62 seconds