Calculus help!!!!?

a.
v'(t) = g
dv/dt = g
dv = g dt
∫dv = ∫g dt
v = gt + C, where C is a constant

Then, when t = 0, v = 34 m/s
34 = g*0 + C
C = 34
g = -9.8 m/s

Hence, v = -9.8t + 34 (m/s)


b.
h = height of the object

dh/dt = v
dh = v dt
dh = (-9.8t + 34) dt
h = -4.9t²+ 34t + C₁, where C₁ is a constant.

When t = 0 s, h = 0 m
0 = -4.9(0)²+ 34(0) + C₁
C₁ = 0
Hence, h = -4.9t²+ 34t


c.
h = -4.9t²+ 34t
dh/dt = -9.8t + 34
d²h/dt² = -9.8

When t = 34/9.8 s = 3.5 s :
h = -4.9(34/9.8)²+ 34(34/9.8) = 59 m
dh/dt = 0
d²h/dt² = -9.8 < 0

Hence, highest point at t = 3.5 s
and height of the highest point = 59 m


d.
h = -4.9t²+ 34t

When h = 0 :
-4.9t²+ 34t = 0
(-4.9t+ 34) t = 0
t = 34/4.9 = 6.9 s or t = 0 (rejected for this is the initial time)

Time taken = 6.9 s

The vertical projectile motion is given by the formula:

h(t) = ½ a t² + v₀t + h₀
where:
a = acceleration due to the gravity
v₀ = initial velocity
h₀ = initial height
So in this case:
a = g = -9.8 m/s² => notice that the negative sign is for direction of the force of gravity which is down
v₀ = 34 m/s
h₀ = 0 => ground
thus:

h(t) = - 4.9 t² + 34t

a)

v(t) = d(h)/dt = - 9.8 t + 34

b) Already found above:

h(t) = - 4.9 t² + 34t

c) at the highest point the velocity is zero:

-9.8t + 34 = 0
9.8t = 34
t = 34/9.8
t ≈ 3.5 seconds

d) the object hits the ground when the height is zero:

h(t) = - 4.9 t² + 34t = 0

-t(4.9t - 34) = 0
t = 0, 34/4.9 => time t = 0 is prior to popping the softball in the air, thus ignore for this part:
t = 34/4.9 seconds
t ≈ 6.9 seconds

I hope this helps.