# Thread: a rock is thrown straight upward with an initial velocity of 50 m/s from a point 100

1. ## a rock is thrown straight upward with an initial velocity of 50 m/s from a point 100

a rock is thrown straight upward with an initial velocity of 50 m/s from a point 100 m above the ground. If the acceleration of the rock at any time t is a= -10 m/$\displaystyle s^2$ , what is the maximum height of the rock (in meters)?

here's my attempt:
a(t)= -10t

take anti-derivative

$\displaystyle v(t)= -5t^2+c$

v(0)= -5(0)^2+c
50= c

$\displaystyle v(t)= -5t^2+50$

$\displaystyle s(t)= \frac{-5}{3}t^3+50t+100$

then i dont know what to do from here

2. Originally Posted by yoman360
a rock is thrown straight upward with an initial velocity of 50 m/s from a point 100 m above the ground. If the acceleration of the rock at any time t is a= -10 m/$\displaystyle s^2$ , what is the maximum height of the rock (in meters)?
a(t) = -10

$\displaystyle \int a(t) dt = v(t) + C = -10t + C$

Initial velocity is v(0), which is 50

$\displaystyle v(0) = -10 \cdot 0 + C \implies v(0) = C = 50$

$\displaystyle v(t) = -10t + 50$

The maximum height of the rock occurs when the velocity is 0. (Think about it, that is the time when the rock is neither moving up nor down)

$\displaystyle v(t) = -10t + 50 = 0$ at max height

$\displaystyle t = 5$

Now you know when the maximum occurs, at t = 5

$\displaystyle \int v(t) dt = s(t) + C = -5t^2 + 50t + C$

The initial position is 100. So s(0) = C = 100

$\displaystyle s(t) = -5t^2 + 50t + 100$

Maximum = s(5)

$\displaystyle s(5) = -125 + 250 + 100 = 225$