Respuesta :
Answer:
Initial Value Problem: [tex]\frac{d^2 x}{dt^2} = -32, x(0) = 200, \frac{dx}{dt}(0) = 40[/tex]
[tex]x(t) = -16t^2 + 40t +200[/tex]
Step-by-step explanation:
The ball is thrown vertically downward, this means that acceleration due to gravity, [tex]g = \frac{dx^{2} }{dt^{2} } = - 32 ft/s^2[/tex]
The height of the ball at time, t = 0 at the top of the building will be: [tex]x(0) = 200 ft[/tex]
The velocity at which the ball is thrown from the top of the building, [tex]\frac{dx}{dt} (0)= 40 ft/s[/tex]
Therefore the initial value problem is written below:
[tex]\frac{d^2 x}{dt^2} = -32, x(0) = 200, \frac{dx}{dt}(0) = 40[/tex]
Let us solve for x(t)
[tex]\frac{d^2 x}{dt^2} = -32\\d(\frac{dx}{dt} )= -32 dt\\[/tex]
Integrate both sides
[tex]\frac{dx}{dt} = -32t + k_1\\\frac{dx}{dt} (0) = 40\\40 = -32(0) + k_1\\k_1 = 0\\\frac{dx}{dt} = -32t + 40[/tex]
Integrate both sides
[tex]x(t) = -16t^2 + 40t + k_2\\x(0) = 200\\200 = -16(0) + 40(0) + k_2\\k_2 = 200\\x(t) = -16t^2 + 40t +200[/tex]
