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A brick is thrown upward from the top of a building at an angle of 25degrees to the horizontal and with an initial velocity of 15m/s. If the brick is in flight for 3 seconds, how tall is the building?
 one year ago
 one year ago
Worst worded question to date: A brick is thrown upward from the top of a building at an angle of 25degrees to the horizontal and with an initial velocity of 15m/s. If the brick is in flight for 3 seconds, how tall is the building?
 one year ago
 one year ago

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amistre64Best ResponseYou've already chosen the best response.0
It took me a little bit to assume that the 3 seconds in flight was a terminanting time, and not just a random interval :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
dw:1346854922454:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
\[4.9(3)^2+15(3)sin(25^o)+h_o=0\]
 one year ago

ash2326Best ResponseYou've already chosen the best response.2
Could I suggest something? @amistre64 sir
 one year ago

ash2326Best ResponseYou've already chosen the best response.2
only the velocity in vertical direction is of use here \[V_v= 15 \sin 25 \] now let's find the time it takes the velocity in vertical direction to get to 0 and the distance it coversdw:1346855419638:dw let the height be h1 and time be t1 here acceleration due to gravity=10 m/s^2 in downwards direction so 10 \[vu=at\] \[0V_v=10*t_1\] we could find t_1 from here \[v^2u^2=2as\] \[0(V_V)^2=2*10*h_1\] h1 from here now the remaining time it takes to reach ground 3t1 using this we could find h2 building's height is h2h1
 one year ago

ash2326Best ResponseYou've already chosen the best response.2
for h2 \[s=ut+1/2 at^2\] \[h_2=0\times (3t_1)+\frac 12 (10) (3t_1)^2\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
while i can understand the concepts you presented, it does seem like an awful lot of extra workings. I am a mathematician at heart, and by default very lazy :)
 one year ago
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