A science project studying catapults sent a projectile into the air with an initial velocity of 24.5 m/s. The formula for distance (s) in meters with respect to time in seconds is
s = -4.9t2 + 24.5t.
a)Find the time that this projectile would appear to have the maximum distance above ground.
B)Find the slope of the tangent at that point using the limit as h approaches 0 of f(x+h) - f(x) over h
2 answers
The maximum height is the VERTEX. You have the equation, and just need to use grade 10 math : ) for the vertex.
actually I think the 24.5 is the initial velocity component up and 4.9 is (1/2)g downward
If
g = -9.8 m/s^2
then
v = Vi - g t where Vi is initial speed up
and height your s, is height up above starting point
s = Vi t - (1/2) g t^2
s = 24.5 t - 4.9 t^2
Now I could tell you easily when it is at the top. It is when v = 0
so 9.8 t =24.5
but yu are supposed to find the vertex of the parabola I think
4.9 t^2 -24.5 t = -s
t^2 - 5 t = (1/4.9)- s
t^2 - 5 t + 2.5^2 = -(1/4.9)s + 6.25
(t-2.5)^2 = (1/4.9)s + 30.625/4.9 = -(1/4.9)(s-30.625)
so
top at t = 2.5 seconds and s = 30.625 meters
now you know that ds/dt = 0 = v at the top but anyway
you know
s(t+h) = -4.9(t+h)^2 + 24.5 (t+h) = -4.9(t^2+2th+h^2) + 24.5 t + 24.5 h
= -4.9 t^2 - 9.8 t h -4.9 h^2+ 24.5 t + 24.5 h
s(t) = -4.9 t^2 +24.5 t
s(t+h) - s(t) = -9.8 th -4.9h^2 +24.5 h
divide by h to get ds/dt (which I call v :)
ds/dt = -9.8 t -4.9h +24.5 = 24.5 - 9.8 t when h-->0
so I say the speed up = 24.5 - 9.8 t
now when t = 2.5, the top
ds/dt = 24.5 - 9.8(2.5) = 24.5 - 24.5 = 0 Whew !
If
g = -9.8 m/s^2
then
v = Vi - g t where Vi is initial speed up
and height your s, is height up above starting point
s = Vi t - (1/2) g t^2
s = 24.5 t - 4.9 t^2
Now I could tell you easily when it is at the top. It is when v = 0
so 9.8 t =24.5
but yu are supposed to find the vertex of the parabola I think
4.9 t^2 -24.5 t = -s
t^2 - 5 t = (1/4.9)- s
t^2 - 5 t + 2.5^2 = -(1/4.9)s + 6.25
(t-2.5)^2 = (1/4.9)s + 30.625/4.9 = -(1/4.9)(s-30.625)
so
top at t = 2.5 seconds and s = 30.625 meters
now you know that ds/dt = 0 = v at the top but anyway
you know
s(t+h) = -4.9(t+h)^2 + 24.5 (t+h) = -4.9(t^2+2th+h^2) + 24.5 t + 24.5 h
= -4.9 t^2 - 9.8 t h -4.9 h^2+ 24.5 t + 24.5 h
s(t) = -4.9 t^2 +24.5 t
s(t+h) - s(t) = -9.8 th -4.9h^2 +24.5 h
divide by h to get ds/dt (which I call v :)
ds/dt = -9.8 t -4.9h +24.5 = 24.5 - 9.8 t when h-->0
so I say the speed up = 24.5 - 9.8 t
now when t = 2.5, the top
ds/dt = 24.5 - 9.8(2.5) = 24.5 - 24.5 = 0 Whew !
1 answer