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Magician
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#1
Report Thread starter 14 years ago
#1
Stone is projected from a point O on a cliff with speed of 20ms^-1 at an angle of elevation of 30degrees. T seconds later, the angle of depression of the stone from O is 45degrees. Find the value of T.


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Gaz031
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#2
Report 14 years ago
#2
Edit: It seems your question refers to displacement rather than velocity.
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Magician
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#3
Report Thread starter 14 years ago
#3
(Original post by Gaz031)
After T seconds let the speed of the particle be V.

Resolving horizontally after the T seconds, the horizontal velocity is unchanged:
20cos30 = Vcos45
V = 20cos30/cos45

Resolving vertically after the T seconds:
u=20sin30=10, a=-9.8, t=T, v=-(20cos30/cos45)sin45.
v = u + at, solve to give T.
This does not match the answer given at the back of the book.
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Fermat
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#4
Report 14 years ago
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After T secs, horizontal distance travelled is H = 20cos30*T
H = 10rt(3)T
=========

After T secs, vertical distance travelled is S = 20sin30*T - ½gT²
S = 10T - 4.9T²
===========

Since angle of depression of particle, from the point O, is 45 deg., then the distance H = the distance S.
Visualise this by drawing a parabolic curve.Let the particle be at the point P after T secs. Then the particle will be x metres horizontally away from O, and also x metres vertically below O, so that its angle of depression is 45 deg.

S = H

SInce S is a negative quantity, change the sign of S.

4.9T² - 10T = 10rt(3)T
4.9T² - (10 + 10rt(3))T = 0
T = 0, T = (10 + 10rt(3))/4.9
T = 5.576
T = 5.6 secs
=========
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Magician
Badges: 15
#5
Report Thread starter 14 years ago
#5
(Original post by Fermat)
After T secs, horizontal distance travelled is H = 20cos30*T
H = 10rt(3)T
=========

After T secs, vertical distance travelled is S = 20sin30*T - ½gT²
S = 10T - 4.9T²
===========

Since angle of depression of particle, from the point O, is 45 deg., then the distance H = the distance S.
Visualise this by drawing a parabolic curve.Let the particle be at the point P after T secs. Then the particle will be x metres horizontally away from O, and also x metres vertically below O, so that its angle of depression is 45 deg.

S = H

SInce S is a negative quantity, change the sign of S.

4.9T² - 10T = 10rt(3)T
4.9T² - (10 + 10rt(3))T = 0
T = 0, T = (10 + 10rt(3))/4.9
T = 5.576
T = 5.6 secs
=========
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