An arrow is shot at 30o above the horizontal. Its velocity is 49 m/s, and it hits the target. a. What is the maximum height the arrow will...

Hello!

I think we ignore air resistance. Then the only force acting on a flying arrow is the gravity force. It is acting downwards and gives an arrow constant acceleration of g (also downwards). Therefore an arrow's height above the initial point is

` V_0*sin(alpha)*t-(g*t^2)/2, `

where `V_0` is the initial speed and `alpha` is the angle above horizontal.

This is an equation of a parabola, and its vertex is at `t=V_0*sin(alpha)/(g).` The height at that moment is a maximum and it is

`(V_0*sin(alpha))^2/(2g) = (49^2*1/4)/(2*9.8)=10*49/16 approx 30.6(m).`

This is the answer for a.

b. An arrow will have the same (zero) height when the formula gives zero for positive t, i.e. at `t=2*V_0*sin(alpha)/g.`

Note that horizontally an arrow have a constant speed `V_0*cos(alpha).` So the distance will be

`2*V_0^2*sin(alpha)*cos(alpha)/g approx 212 (m).`