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...
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.
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