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

first, find the vertical speed of 15m/s at 15° to the horizontal.
V0 = 15*sin15 = 3.88m/s
time taken, t = 3.2s.

taking upwards as positive,
using eqn, X-X0=1/2 at^2 + V0t
X-X0 = 1/2(-9.81)(3.2)^2 + 3.88(3.2)
= -50.23 + 12.42
= -37.81m

The stone will be displaced 37.81m vertically after 3.2s. Therefore, the building is 37.81m high.

Answer 2

X-X0=1/2 at^2 + V0t

Use this one but rearrange it

1/2 at^2 + V0t + X0 = X

With X0 being your unknown

They give you the final position which is 0 at t=3.2
Assuming the building is on earth, they give you acceleration due to gravity, ~-9.8m/s^2

They give you a velocity of 15 m/s at an angle of 15 degrees
Since horizontal component of this velocity does not affect the flight time, you need to find just the vertical component.
Vy = 15 sin 15

put it all together..

-4.9t^2 + 15 sin(15)*t + H = X

Since t = 3.2 when the ball hits the ground (height is 0)
-4.9(3.2)^2 + 15sin(15)*3.2 + H =0
-37.75223 + H = 0
-H = -37.7523
H = 37.75

Answer 3

I got a different answer from yours( anyone else who can prove who is right or wrong will be appreciated.)

(assuming there is no friction and the free fall is 10m/s/s.)
This is what I did:
If the speed is 15m/s and it's moving up at 15degree to horizontal, then the upward rate of change in distance initially will be:
sin 15degree= upward rate of change in distance/ 15
upward rate of change in distance= (sin 15)* 15
= it's around 3.88m/s
(if u don't get this, draw a right-angle triangle with 15 as hypotenus and meet with horizontal at 15degree, then just find the length of the vertical side.)

Since the brick is moving up then will fall again, we need to find the time it takes to travel the distance that is not the length of the building and minus it from the total time.

The upward speed is 3.88m/s. The brick comes to a mid-air stop in:
v= u+at
0= 3.88- 10t ( it's negative 10 as the gravity is acting against the motion of the brick.)
t= 3.88/10= 0.388sec
The brick will take another 0.388sec to cover the distance it moved up so the time taken to cover the distance which is above the building is 0.388x2= 0.776 sec

3.2sec-0.776sec= 2.424sec (time taken to fall the length of the building.)
The speed when it reaches back the building is the same as the speed it started off which is 3.88m/s.
Now we put all the information into the formula:

s= ut+ (1/2)(a)(t^2)
s= (3.88)(2.424)+ (1/2)(10)(2.424^2)
s= 38.78m

It's around your answer but still not got quite a large difference. Hope my idea won't confuse you even more.

(Meaning of the symbols:
s
v
u
a
t