CBSE Class 12-science Answered
A person standing on the top of the 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 15. Ft/ s , how short will the packet fall.
Asked by nehagoyalarmy | 16 Jun, 2018, 11:24: AM
Expert Answer
As shown in figure, let a person A is at a height 171 feet from ground level.
Person A throwing a packet to another person B, who is 228 feet horizontally away.
The straight line distance from A to B is 
feet.
feet.Packet is thrown along the direction AB with speed 15 ft/s. If we resolve this speed into vertical and horizontal components,
we get vertical component as 9 ft/s ( 15 sinθ = 15×171/285 = 9 ft/s) and horizontal component as 12 ft/s ( 15 cosθ = 15×228/285 = 12 ft/s )
The time to reach the ground for the packet is decided by gravity.
Hence we will find the time taken to reach the ground using vertical component.
we need to use the equation of motion, " S = ut+(1/2)gt2 "
171 = 9×t+(1/2)×9.8×3.281×t2 = 9×t+16.08×t2 ...................(1)
(acceleration due to gravity is ft/s2. 1 m = 3.281 ft)
solving the quadratic equation (1) for t, we get t = 3 s.
Hence horizontal distance travelled by the packet = 12×3 = 36 ft.
Packet will fall at a distance 36 ft from O, short of 192 ft to B ( 228-36 = 192 ft )
Answered by Thiyagarajan K | 16 Jun, 2018, 03:28: PM
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