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Question
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. Horizontal away. If he throws the packet directly aiming at the friend with speed of 15 ft/s. How short 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. Horizontal away. If he throws the packet directly aiming at the friend with speed of 15 ft/s. How short the packet fall??
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Solution
Since the person on the cliff throws the stone directly towrds the person on ground, then while throwing he makes an angle θ (say) with the vertical cliff.
=> tanθ= 228/171 = 4/3
=> θ= 53°
taking the standard x-y coordinate system,
in 'y' direction,
inital velocity = 15 cos θ = 15 cos 53° = 9 ft/s
v² = u² + 2 g h
where g = 32 ft/s , h = 171 ft , u = 9 ft/s
v = 105ft/s
=> v = u + gt => 105 = 9 + 32t
=> t = 3 s
now in 'x' direction, initial velocity = 15 sin 53 = 12 ft/s
distance travelled in 3 s =∆12 × 3 = 36 ft
distance throgh which it falls short = 228 - 36 =192 ft
=> tanθ= 228/171 = 4/3
=> θ= 53°
taking the standard x-y coordinate system,
in 'y' direction,
inital velocity = 15 cos θ = 15 cos 53° = 9 ft/s
v² = u² + 2 g h
where g = 32 ft/s , h = 171 ft , u = 9 ft/s
v = 105ft/s
=> v = u + gt => 105 = 9 + 32t
=> t = 3 s
now in 'x' direction, initial velocity = 15 sin 53 = 12 ft/s
distance travelled in 3 s =∆12 × 3 = 36 ft
distance throgh which it falls short = 228 - 36 =192 ft
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