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amistre64
Group Title
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?
 2 years ago
 2 years ago
amistre64 Group Title
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?
 2 years ago
 2 years ago

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amistre64 Group TitleBest 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 :)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1346854922454:dw
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[4.9(3)^2+15(3)sin(25^o)+h_o=0\]
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Could I suggest something? @amistre64 sir
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
of course
 2 years ago

ash2326 Group TitleBest 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
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
for h2 \[s=ut+1/2 at^2\] \[h_2=0\times (3t_1)+\frac 12 (10) (3t_1)^2\]
 2 years ago

amistre64 Group TitleBest 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 :)
 2 years ago
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