Hello,
I have one more problem today.
A cannister is dropped from a helicopter 500m above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100m/s. Will it burst?
I thought I had this one figured out finally but my answer doesn't match that of the book. (Will it burst? The books answer, simply: no.)
I know that the pull of gravity is -9.8m/s^2
a(t) = -9.8t
antiderivative of a(t) is v(t)
v(t) = -4.9t^2 + C
antiderivative of v(t) is s(t)
s(t) = (-4.9/3)t^3 + Ct + D
D is 500m since that is where the cannister began falling from.
s(t) = (-4.9/3)t^3 + Ct + 500
I need to find the velocity at the point in time where s(t), the position, is 0.
0 = (-4.9/3)t^3 + Ct + 500
I found an equation in the book...
s = 1/2at^2 + v0t + s0
where v0 and s0 mean the initial velocity and position.
If I set v0 to 0, s0 to 500, a to -9.8, and s to 0 (the point where it hits the ground), I end up with: t = 10.1015s
Plug that into my s(t) to solve for C:
0 = -4.9/3(10.1015)^3 + 10.1015C + 500
C = 117.3
Replace C to solve for the velocity at t=0:
v(10.1015) = -4.9(10.1015)^2 + 117.3
v = -382.829m/s
But that velocity is MUCH faster than 100m/s, so the can should burst. Does anyone know where I went wrong above, or am I doing the entire problem incorrectly?
I have one more problem today.
A cannister is dropped from a helicopter 500m above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100m/s. Will it burst?
I thought I had this one figured out finally but my answer doesn't match that of the book. (Will it burst? The books answer, simply: no.)
I know that the pull of gravity is -9.8m/s^2
a(t) = -9.8t
antiderivative of a(t) is v(t)
v(t) = -4.9t^2 + C
antiderivative of v(t) is s(t)
s(t) = (-4.9/3)t^3 + Ct + D
D is 500m since that is where the cannister began falling from.
s(t) = (-4.9/3)t^3 + Ct + 500
I need to find the velocity at the point in time where s(t), the position, is 0.
0 = (-4.9/3)t^3 + Ct + 500
I found an equation in the book...
s = 1/2at^2 + v0t + s0
where v0 and s0 mean the initial velocity and position.
If I set v0 to 0, s0 to 500, a to -9.8, and s to 0 (the point where it hits the ground), I end up with: t = 10.1015s
Plug that into my s(t) to solve for C:
0 = -4.9/3(10.1015)^3 + 10.1015C + 500
C = 117.3
Replace C to solve for the velocity at t=0:
v(10.1015) = -4.9(10.1015)^2 + 117.3
v = -382.829m/s
But that velocity is MUCH faster than 100m/s, so the can should burst. Does anyone know where I went wrong above, or am I doing the entire problem incorrectly?