Question

A snowball is thrown straight upward from the top of a 200 meter tall building with a velocity of 30 m/s. How long does it take to reach its highest point?

Answer 1

`t=v_o/g = (30m//s)/(9.8m//s^2)=3.06s`

Explanation:

One way to look at this problem is to think about what happens when the snowball reaches its highest point. That is the point when it is no longer moving upwards and then begins its decent. That means that at the highest point the velocity is zero. I'm also assuming that the question is asking us to find the time that it takes to reach this point by asking "how long" it takes.

We know that under gravity, the acceleration, `g`, is constant, and we know the initial velocity, `v_o`, of the snowball. So we can express the velocity as a function of time as:

`v(t) = v_o - g t`

As discussed we are trying to find the time at which `v(t)=0` therefore we can solve for `t` in

`0=v_o-g t`

which yields

`t=v_o/g = (30m//s)/(9.8m//s^2)=3.06s`

If we want to find the distance, there is a simple way to do that. If we look at the equation for `v(t)` we notice that it changes linearly with time.

`v(t) vs. t:`

graph{-9.8*x+30 [-0.2, 3.1, -4, 30]}

Therefore, during its ascent, the average velocity is just half of the initial velocity. We can multiply this by the time we just obtained to get the height above the building the ball travels:

`h = bar(v) * t = 15m//s * 3.06s = 46m`

Therefore the maximum height above the ground the snowball reaches is `246m`