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Grade: upto college level
        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?
5 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
471 Points
							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

						
5 years ago
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