antiderivative word problem

Question

A ball is shot from the ground straight up into the air with initial velocity of 49 ft/sec. Assuming that the air resistance can be ignored, how high does it go?

Answer 1

x(0)=0

v(0) = x'(0) = 49

a(0) = v'(0) = x''(0) = -32

a(t) = -32

v(t) = -32t + C

v(0) = 49 ---> C=49

So v(t) = -32t + 49

x(t) = -16t^2 + 49t + C

x(0) = 0 --->C=0

the position function is x(t) = -16t^2 + 49t

velocity function is v(t) = x'(t) = -32t + 49

velocity is zero when -32t + 49 = 0

t = 49/32

max height is x(49/32) = -16 ( 49/32)^2 + 49 ( 49/32)

= [-16 ( 49 * 49) + (49*49*32)]/ (32*32)

= (49*49) [ -16 + 32] / (32*32)

= (49*49) (16) / (32*32)

= (49*49) / (32*2)

= (49*49)/64

= (50-1)(50-1)/64

= (2500 - 100 + 1)/64

= 2401/64