An Integral Problem- Kindly solve with solutions. Thank you.?

Question

A stone is thrown vertically upward from the top of a house. If it has reached its maximum height of 30 ft after 1 second. Find the height of the house- Assuming the only force acting is acceleration due to gravity (-32 ft/sec^2

Answer 1

The integral of acceleration (-32ft/sec^2) is velocity:

v(t) = -32t + c

At its maximum height, velocity is zero:

v(1) = -32 + c
0 = -32 + c
c = 32

v(t) = -32t + 32

Height is the integral of velocity:

h(t) = -16t^2 + 32t + c

At one second, its height is 30 feet:

h(1) = -16 + 32 + c
h(1) = 16 + c
30 = 16 + c
c = 14

The position at time t is thus:

h(t) = -16t^2 + 32t + 14

The position at time t=0 is the top of the house:

h(0) = height of house
-16(0^2) + 32*0 + 14 = height
14 = height

Answer 2

v=u+at v=0 a= -32ft/s^2 t= 1s u=32ft/s v^2=u^2+2as 0=32^2 - 2*32*s s=(32*32)/2*32 =sixteen ft from the right of the living house. height of the living house = optimal height - distance from the right of the living house = 30 - sixteen = 14 ft

Answer 3

Why do you need to ask a question like this in this column?ask a question which you truly need to know and but not testing people.

Answer 4

seems to be a complex problem