Question

A projectile is fired from the edge of a 240m cliff with an initial speed of 65m/s at an angle of 45degrees above the horizontal. how much time until projectile hits the ground below?

Answer #1

Height of the cliff H = 240 m

Initial speed v = 65 m/s

Angle = 45
^{o}

Required time T = ?

Accleration a = -g = -9.8m/s ^{2}

Time taken to reach maximum height t = v sin 45 / g

= 65 sin 45 / 9.8

= 4.689 s

Maximum height above the top of the cliff h = (v sin 45)
^{2} / 2g

= (65 sin 45) ^{2} /(2x9.8)

= 107.78 m

Maximum height from ground h ' = H + h = 240 m + 107.78 m = 347.78 m

Time taken to reach the ground from maximum height t ' = [2h ' / g ]

t ' = [(2x347.78)/9.8]

= 8.424 s

So, required time T = t + t ' = 4.689 s + 8.424 s

= 13.113 s

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