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A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall?

A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall?

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
6 years ago
Sol. tan θ = 171/228 ⇒ θ = tan–1 (171/228) The motion of projectile (i.e. the packed) is from A. Taken reference axis at A. ∴ θ = –37° as u is below x-axis. u = 15 ft/s, g = 32.2 ft/s2, y = –171 ft y = x tan θ – (x^2 g sec^2⁡θ)/(2u^2 ) ∴ –171 = –x (0.7536) – (x^2 g (1.5568))/(2(225)) ⇒ 0.1125x2 + 0.7536 x – 171 = 0 x = 35.78 ft (can be calculated) Horizontal range covered by the packet is 35.78 ft. So, the packet will fall 228 – 35.78 = 192 ft short of his friend

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