A projectile takes off with an initial velocity of 10m/s at an angle of elevation of 45° .it is just able to clear two hurdles of height 2 m each,separated from each other by a distance d.calculate d.at what distance from the point of projection is the first hurdle placed? Take g= 10 m/s^2

Question

Arun · Answer

Use the trajectory equation:
y = h + x·tanΘ - g·x² / (2v²·cos²Θ)
where y = height at x-value of interest = 2 m
and h = initial height = 0 m
and x = range of interest = ???
and Θ = launch angle = 45º
and v = launch velocity = 10 m/s

Dropping units for ease,
2 = 0 + xtan45 - 9.8x²/(2(10)²cos²45) = x - 0.098x²
This quadratic has roots at
x = 2.73 m ← first hurdle
and x = 7.47 m ← second hurdle

d = 7.47m - 2.73m = 4.74 m ◄

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A projectile takes off with an initial velocity of 10m/s at an angle of elevation of 45° .it is just able to clear two hurdles of height 2 m each,separated from each other by a distance d.calculate d.at what distance from the point of projection is the first hurdle placed? Take g= 10 m/s^2

A projectile takes off with an initial velocity of 10m/s at an angle of elevation of 45° .it is just able to clear two hurdles of height 2 m each,separated from each other by a distance d.calculate d.at what distance from the point of projection is the first hurdle placed? Take g= 10 m/s^2

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A projectile takes off with an initial velocity of 10m/s at an angle of elevation of 45° .it is just able to clear two hurdles of height 2 m each,separated from each other by a distance d.calculate d.at what distance from the point of projection is the first hurdle placed? Take g= 10 m/s^2

A projectile takes off with an initial velocity of 10m/s at an angle of elevation of 45° .it is just able to clear two hurdles of height 2 m each,separated from each other by a distance d.calculate d.at what distance from the point of projection is the first hurdle placed? Take g= 10 m/s^2

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