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A person standing on top of a cliff 171ft high has to throw a packet to his friends standing on the ground 228ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/sec, how short will the packet fall? Ans: 192 ft Can You Please Explain The Solution To Me I Am A Xth Standard Student ?

A person standing on top of a cliff 171ft high has to throw a packet to his friends standing on the ground 228ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/sec, how short will the packet fall?
Ans: 192 ft

Can You Please Explain The Solution To Me I Am A Xth Standard Student ?

Grade:10

1 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

g=32ft/sec2,horizontal component of velocity is constant throughout the motion, and equal to (15*228)/√(2282+1712)=v1

initial vertical velocity component =v2=(15*171)/√(2282+1712),let time taken to hit ground be 't',171=(v2*t)+(g/2)t2,even though this is a quadratic equation, the choice of the correct root tc will be obvious. horizontal distance travelled= (v1*tc)

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