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 = u2sin2θ/g —–1
T = 2usinθ/g —–2
now expanding the equation for range R , we get R = 2u2sinθcosθ/g —-3
dividing equation 3 by equation 2
we get [2u2sinθcosθ/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.
Tr = 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
hmax = (v2 – (u sinθ)2)/2 g
hmax = (02 – 252)/(2 × (-10)) = 31.25 m