# An arrow is fired straight upwards at a speed of 50.0m/s from the top of a building 100m above the ground. How long does it take to strike the ground at the base of the building? What was the...

An arrow is fired straight upwards at a speed of 50.0m/s from the top of a building 100m above the ground. How long does it take to strike the ground at the base of the building? What was the maximum height reached? What was the final speed before impact with the ground?

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Hello!

Using Newton's Second Law, one can obtain the known formulas for such a movement:

`V(t)=V_0-g t,` `H(t)=H_0+V_0t-g t^2/2,`

where `t` is the time from the start, `V_0` is the initial speed, `H_0` is the initial height and `g approx 10m/s^2` is the gravity acceleration. `V` is the speed of an arrow as a function of time, `H` is its height.

The time `t_1` until an arrow strikes the ground must be positive and satisfy the equation `H(t_1)=0` (ground level is assumed zero). This is a quadratic equation. In numbers it is:

`100+50t-5t^2=0,` or `t^2-10t-20=0.`

So `t_1=5+sqrt(25+20) approx 11.7(s).` I used the quadratic formula here. The solution before the root is negative.

Next, the maximum height is when an arrow finishes its rise and will begin to fall, i.e. when `V(t)=0.` This is `t=V_0/(g)=5(s).` The height above the ground is `H(5)=100+250-125=225(m).`

And the final speed is the speed at `t_1,` `V(t_1)=50-10*11.7=-67(m/s).` This means `67m/s` downwards.

So the answers are: an arrow strikes the ground after about **11.7 s**, its speed before that moment will be about **67 m/s** downwards, and the maximum height above the ground will be about **225 m**.