Two particles were projected one by one with the same initial velocity from the same point on level ground. They follow the same parabolic trajectory and are found to be in the same horizontal level, separated by a distance of 1 m, 2 s after the second particle was projected. Assume that the horizontal component of their velocities is 0.5 m s $^{- 1}$ . Find
a. the horizontal range of the parabolic path.
b. the maximum height for the parabolic path.

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Solution

Distance travelled by 2nd particle in $2 s$ = $0.5 \times 2 = 1 m$
Horizontal range $= 1 + 1 + 1 = 3 m$
Flight time $= 4 + 2 = 6 s$
$6 = \frac{2 u sin θ}{g} u sin θ = 30$
$H = \frac{u^{2} {sin}^{2} θ}{2 g} = \frac{900}{2 \times 10} = 45$

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Distance travelled by 2nd particle in 2s = 0.5×2=1mHorizontal range =1+1+1=3mFlight time =4+2=6s6=2usinθgusinθ=30H=u2sin2θ2g=9002×10=45

Distance travelled by 2nd particle in $2 s$ = $0.5 \times 2 = 1 m$
Horizontal range $= 1 + 1 + 1 = 3 m$
Flight time $= 4 + 2 = 6 s$
$6 = \frac{2 u sin θ}{g} u sin θ = 30$
$H = \frac{u^{2} {sin}^{2} θ}{2 g} = \frac{900}{2 \times 10} = 45$