differential equation problem with newton law?

Question

A stone was thrown into air from a roof of a building 100fts. high with the speed of 20ft/s and 30 angle( from the ground level) while neglectin air resistance find:
a) the system of differential equation describing the motion of the stone thogether with the corresponding initial conditions,
b) the solution to the system in part a
c) stones highest altitude.
d) the time needed for the stone to reach the ground,
e) stone's position ( from the building) when it hits the ground,
stones velocity( its a vector!) and speed when it hits the ground
exact result
please can you show all the steps
thank you!

Answer 1

I am very surprised this problem has not been helped yet.
Let us use "_y" as notation of sub-y.
(a) a_y = dv_y/dt = -mg, with v_0y = 10ft/s, v_0x = 10*sqrt(3) ft/s, s_0y = 100fts, and s_0x = 0fts.
(b) integrate it once:
ds_y/dt = v_y = -mgt + 10, and ds_x/dt = v_x = 10*sqrt(3)
integrate it once more:
s_y = -0.5mgt^2 + 10t + 100, and s_x = 10*sqrt(3)t
(c) s_y = -0.5mgt^2 + 10t + 100
= - (sqrt(0.5mg)*t - 5/sqrt(0.5mg))^2 + 100 + 50/mg
Hence stone's highest altitude is (100 + 50/mg)fts.
(d) let s_y = 0 = -0.5mgt^2 + 10t + 100
Solve this quadratic we have:
t = {-10 + sqrt(100 + 200mg)}/mg
(e) s_x = 10*sqrt(3)t
= 10*sqrt(3)*{sqrt(100 + 200mg) - 10}/mg
(v_x, v_y) = (10*sqrt(3), - sqrt(100 + 200mg)
and speed = sqrt((v_x)^2 + (v_y)^2) = sqrt(400 + 200mg)
I just try to help you--you can fill in the numbers.