## amistre64 2 years 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?

1. amistre64

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. amistre64

|dw:1346854922454:dw|

3. amistre64

$-4.9(3)^2+15(3)sin(25^o)+h_o=0$

4. ash2326

Could I suggest something? @amistre64 sir

5. amistre64

of course

6. ash2326

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 covers|dw: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 $v-u=at$ $0-V_v=-10*t_1$ we could find t_1 from here $v^2-u^2=2as$ $0-(V_V)^2=2*-10*h_1$ h1 from here now the remaining time it takes to reach ground 3-t1 using this we could find h2 building's height is h2-h1

7. ash2326

for h2 $s=ut+1/2 at^2$ $h_2=0\times (3-t_1)+\frac 12 (-10) (3-t_1)^2$

8. amistre64

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 :)

9. ash2326

:) he he