A brick is thrown upward from the top of a building at an angle of 15° to the horizontal and with an initial s

Question

A brick is thrown upward from the top of a building at an angle of 15° to the horizontal and with an initial speed of 15 m/s. If the brick is in flight for 3.2 s, how tall is the building?
____m

I know the answer is 37.75 but I can't figure out how it is done. Will someone work this out for me and explain what you are doing. And if possible can you use the equations I put below, as those are the ones we use in class.

V=at+V0
(X-X0)=1/2 (V+V0)t
V^2=V0^2 + 2a (X-X0)
X-X0=1/2 at^2 + V0t

You can use the equations one for y and one of x if you need as im not sure how to do it...but any help would be appreciated. thanks

Answer 1

Your question lacks enough information to give a logical answer. Thanks

Answer 2

You can ignore the x component. The formula you use here is y = y0 + V0y*t + 0.5*g*t^2, where y is vertical position, y0 is the original position (the height of the building), V0y is the initial vertical component of velocity, g is acceleration due to gravity (a negative value, -9.8 m/s^2), and t is elapsed time. Since you know that the brick hits the ground at t = 3.2 s, and that corresponds to a final position of y = 0, you should be able to solve for y0.

You need to start by calculating the initial vertical component of velocity. It's V0y = (15 m/s)*sin(15°). Then you solve the equation y = y0 + V0y*t + 0.5*g*t^2 ==> y0 = y - V0y*t - 0.5*g*t^2 = 0 - V0y*3.2 - 0.5*(-9.8)*3.2^2 = V0y*3.2 + 50.176, where you have already calculated V0y.

Answer 3

We know that V in y axis is 15sin15 = 15*0.258819045
= 3.88228568 m/s
then
X-X0=1/2 at^2 + V0t
X-X0=1/2 9.8*3.2*3.2 + (-3.88228568)3.2
X-X0= 50.176 - 12.4233142
X-X0=37.752686
It must have minus on V because V is opposite way to gravity
My answer is not same as your answer because sin15 we used is not the same but my answer is also correct