Respuesta :
Answer:
(a) 25.08 m
(b) 2.05 m
Step-by-step explanation:
Let the height of the building be 'h'.
Given:
Angle of projection is, [tex]\theta=25[/tex]°
Initial speed is, [tex]u=15\ m/s[/tex]
Time of flight is, [tex]t=3.0\ s[/tex]
(a)
Consider the vertical motion of the brick.
Vertical component of initial velocity is given as:
[tex]u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s[/tex]
Vertical displacement of the brick is equal to the height of the building.
So, vertical displacement = [tex]-h[/tex] (Negative sign implies downward motion)
Acceleration is due to gravity in the downward direction. So,
Acceleration is, [tex]g=-9.8\ m/s^2[/tex]
Now, using the following equation of motion;
[tex]-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m[/tex]
Therefore, the building is 25.08 m tall.
(b)
Let the maximum height be 'H'.
At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.
So, [tex]v_y=0\ m/s[/tex]
Now, using the following equation of motion, we have:
[tex]v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m[/tex]
Therefore, the maximum height of the brick is 2.05 m.
