Q.

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 (in ft) will the packet fall? [Use, g = 32 ft/s2]

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answer is 192.

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Detailed Solution

tan⁡θ=171228=34 sy=uyt+12ayt2 +171=+(15sin⁡θ)T+12(+32)T2⇒171=1535T+16T2⇒16T2+9T−171=0⇒T(16T+57)−3(16T+57)=0⇒T=3s       sx=uxt+12axt2⇒ sx=(15cos⁡θ)T=15×45(3)=36ft      x=36ftSo, packet will tall short by ( ) 228 36 = 192 ft
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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 (in ft) will the packet fall? [Use, g = 32 ft/s2]