Question

Am arrow is thrown in the air.its time of flight is 5s and the range is 200m . determine (1) the vertical component of the velocity of projection,(2) the horizontal component,(3) maximum height and(4) the angle made with the horizontal.(g=9.8m/s Square)

Answer 1

"Given data are the time of flight and the range as 5 seconds and 200 metres respectively and we are asked to calculate the vertical component of the velocity denoted as uy which is equal to u sin ᶿ and the horizontal component of the velocity denoted as ux which is equal to u cos ᶿ, and the maximum height and also the angle of projection. By applying equations of motion, we get all these components. For $ux = u cos^{\theta}$, Range = ux x T, where T is the time of flight, therefore, ux = R/T => ux = 200/5 = 40m/s. For $uy = u sin^{\theta}$, we have $v = u sin^{\theta}- aT$where a is the acceleration due to gravity here, to attain range of projectile final velocity v must be zero => v = 0, => 0 = uy – gT. For the vertical component, T = T/2 due to symmetrical flight of upward and downward. => uy = 10 x 2.5 = 25m/s. For the maximum height, uy must be zero therefore$h=[tex]\frac{ 1 }{ 2 }$g(t/2)2=$\frac{1}{2}$x10x(2.5)2=31.25m[/tex]. For the angle of projection, $tan^{\theta} = uy/ux = 25 / 40 = 0.625 =>^{\theta} = tan-1 (0.625) =>^{\theta} = 32 degree$."

Answer 2

Answer: u sin theta= 24.5m/s

u cos theta= 40m/s

H= 30.6m

Explanation: H= u²sin²theta/2g

T= 2u sin theta/g

R= ucostheta × T