**An arrow is thrown in the air. It’s time of flight is 5 s and the range is 200 … with horizontal.(iii) maximum height(IV) horizontal…**

###### Answers:

To determine horizontal component of velocity

as we know

R = u^{2}sin2θ/g —–1

T = 2usinθ/g —–2

now expanding the equation for range R , we get R = 2u^{2}sinθcosθ/g —-3

dividing equation 3 by equation 2

we get [2u^{2}sinθcosθ/g]/[ 2usinθ/g] = R/T

thus we have

ucosθ = R/T

or ucosθ = 200/5 = 40m/s

as we know ucosθ is the horizontal component of velocity when θ is the angle with the horizontal.

T_{r} = T/2 = 5/2 = 2.5 s

V = u + gt

Vertical component of initial velocity = µ sin θ

V = u sin θ + gt

µ sin θ = v – gt

µ sin θ = 0 – (-10) × 2.5

µ sin θ = 25 m/s

vertical component of initial velocity = 25m/s

h_{max} = (v^{2 }– (u sinθ)^{2})/2 g

h_{max} = (0^{2} – 25^{2})/(2 × (-10)) = 31.25 m