An object is shot up into the sky according to s(t)= -16t^2+64t where s(t) = height above the ground in feet at t seconds. Find the a. velocity at 2 seconds b. acceleration at 1 second c. maximum height of the object.
Let me see, I used feet a few years ago, like 60.
v = ds/dt = -32 t + 64
a = -32 ft/second^2 = constant acceleration of gravity by the way
max height when v = 0 (stops going up, not going down yet)
t = 64/32 seconds
Sorry, I'm not understanding c.
They asked for the maximum height
That is when ds/dt = 0
In other words when the speed up stops
So I'm plugging in 64 (which is what I got from putting 0 into v(t) ) into s(t) and my answer should come out to -61440? I'm confused lol
-b/2a = 64/-32 = -2
So find s(t) when t=2
You don't need any calculus for this. Just think back on your Algebra I
The vertex occurs midway between the roots, which are at t=0 and 4
v = ds/dt = -32 t + 64
0 = -32 t + 64
t = 2 not 64
To find the velocity at a specific time, we need to find the derivative of the height function, s(t).
a. The velocity at 2 seconds is found by taking the derivative of s(t). The derivative of -16t^2 + 64t is:
s'(t) = -32t + 64
To find the velocity at t = 2 seconds, substitute t = 2 into the derivative:
s'(2) = -32(2) + 64
s'(2) = -64 + 64
s'(2) = 0
Therefore, the velocity at 2 seconds is 0 feet per second.
b. To find the acceleration at a specific time, we need to find the derivative of the velocity function.
The derivative of s'(t) = -32t + 64 is:
s''(t) = -32
Therefore, the acceleration of the object is -32 feet per second squared.
c. To find the maximum height of the object, we need to find the vertex of the parabolic function s(t) = -16t^2 + 64t. The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula:
t = -b / (2a)
In this case, a = -16 and b = 64, so we can substitute these values into the formula:
t = -64 / (2*(-16))
t = -64 / (-32)
t = 2
Therefore, the maximum height is reached at t = 2 seconds.
To find the height at t = 2 seconds, substitute t = 2 into the height function:
s(2) = -16(2)^2 + 64(2)
s(2) = -16(4) + 128
s(2) = -64 + 128
s(2) = 64
Therefore, the maximum height of the object is 64 feet.
v(t) = -32t+64
a(t) = -32
as with all quadratics, the vertex is at x = -b/2a
s(t)= -16t^2+64t
v(t) = -32t + 64
a(t) = -32
a) find v(2)
b) clearly a constant for the domain of t
c) set v(t) = 0 and solve for t
then sub that into s(t) to find the corresponding height