微積分應用題，請問圈起來的題目怎麼解?

(a)
英制, 重力加速度 = 32 ft / s^2
h = 256 - (1/2)*32*t^2 = 256 - 16*t^2
落地時, h = 0 , 所以:
0 = 256 - 16*t^2
t^2 = 256/16 = 16
t = 4

(b)
tan θ = h / 500 = ( 256 - 16*t^2 ) / 500
等式兩邊對 t 微分, 等式恆成立:
sec^2 θ * dθ/dt = - (16/500)*2t = - 8t / 125
dθ/dt = - cos^2 θ * 8t / 125

When t = 1
h = 256 - 16 = 240
cos^2 θ = 500^2 / ( 500^2 + 240^2 ) = 250000/307600 = 625/769
dθ/dt = - (625/769) * 8/125 = - 40/769

When t = 2
h = 256 - 16*4 = 192
cos^2 θ = 500^2 / ( 500^2 + 192^2 ) = 250000/286864 = 15625/17929
dθ/dt = - (15625/17929) * 8*2/125 = - 2000/17929

Ans:
(a)
h = 256 - 16*t^2
4 秒時落地
(b)
當 t = 1 , dθ/dt = - 40/769
當 t = 2 , dθ/dt = - 2000/17929