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? Give g = 32 ft/s2.

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= 53o

taking the standard x-y coordinate system,

in 'y' direction,
inital velocity = 15 cos θ = 15 cos 53o = 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 (ans.)


To your doubt.The direcyion of g is downwards.We are throwing the stone also in a downward direction.Since both are in downward direction,we take g as positive.