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 ?Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics
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95
Solution :
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Please refer the attachment for the diagram .
From the figure, Tanθ=171/228
θ=Tan⁻¹[171/228]
The motion of projectile is from point A. Take reference axis at a.
θ=37° as u is below x axis.
u=15ft/s
g=32.2ft/s2
y=-171ft
From y= Tanθ-x²gsec²θ/2u²
-171=(tan37°)x- x²x32.2x(1.568)/2x225
-171=0.7536x- 0.1125x²
0.1125x²-0.7536x-171=0
on solving we can get x=35.78ft
Horizontal range covered by packet is 35.8ft .
So the packet will fall =228-35.8=192ft short of his friend.
****************************************************************
Please refer the attachment for the diagram .
From the figure, Tanθ=171/228
θ=Tan⁻¹[171/228]
The motion of projectile is from point A. Take reference axis at a.
θ=37° as u is below x axis.
u=15ft/s
g=32.2ft/s2
y=-171ft
From y= Tanθ-x²gsec²θ/2u²
-171=(tan37°)x- x²x32.2x(1.568)/2x225
-171=0.7536x- 0.1125x²
0.1125x²-0.7536x-171=0
on solving we can get x=35.78ft
Horizontal range covered by packet is 35.8ft .
So the packet will fall =228-35.8=192ft short of his friend.
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Answered by
100
HEY!!
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✔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,
Iin 'y' direction,
✔Inital velocity = 15 cos θ = 15 cos 53° = 9 ft/s
v^2 =u^2 + 2gh where g=32 ft/s, h=171ft, u=9ft/s
v = 105ft/s
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.)
_______________________________
✔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,
Iin 'y' direction,
✔Inital velocity = 15 cos θ = 15 cos 53° = 9 ft/s
v^2 =u^2 + 2gh where g=32 ft/s, h=171ft, u=9ft/s
v = 105ft/s
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.)
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