Respuesta :
Answer:
a) 2.0 s is the time after which the balloon hit the building
a) 49.3 m is the height at which the balloon hits the building
d) 24.8 m/s is the velocity of the balloon at the instant of hitting the building
Explanation:
Given:
initial speed of projection of balloon, [tex]u=40\ m.s^{-1}[/tex]
angle of projection from the horizontal, [tex]\theta=60^{\circ}[/tex]
distance of a tall building from the launch site, [tex]s=40\ m[/tex]
Now we find the horizontal component of velocity :
[tex]u_x=u\cos\theta[/tex]
[tex]u_x=40\cos60[/tex]
[tex]u_x=20\ m.s^{-1}[/tex]
Now the time taken to hit the building:
(since we don't have any acceleration in the horizontal direction )
[tex]t=\frac{s}{u_x}[/tex]
[tex]t=\frac{40}{20}[/tex]
[tex]t=2\ s[/tex]
Now the vertical component of the velocity:
[tex]u_y=u\sin\theta[/tex]
[tex]u_y=40\sin60[/tex]
[tex]u_y=34.64\ m.s^{-1}[/tex]
Time taken to reach the maximum height:
At max. height the final velocity (v=0)
[tex]v_y=u_y-g.t_m[/tex] (-ve, since the direction of velocity is opposite to the gravity)
[tex]0=34.64-10\times t[/tex]
[tex]t=3.464\ s[/tex]
Therefore the the balloon hit the building while ascending in height.
Now the height reached in the time in which the balloon hits the building:
[tex]h=u_y.t-\frac{1}{2} g.t^2[/tex] (-ve, since the direction of velocity is opposite to the gravity)
[tex]h=34.64\times 2-0.5\times 10\times 2^2[/tex]
[tex]h=49.28\ m[/tex]
Speed of the balloon while it hits the building:
- (The horizontal component of the velocity remains constant during the motion)
[tex]u_x=20\ m.s^{-1}[/tex]
- The vertical component of velocity at time t=2 sec:
[tex]v_y'=u_y-g.t[/tex] (-ve, since the direction of velocity is opposite to the gravity)
[tex]v_y'=34.64-10\times 2[/tex]
[tex]v_y'=14.64\ m.s^{-1}[/tex]
Now the net resultant of the two components:
[tex]v=\sqrt{(u_x)^2+(v_y')^2}[/tex]
[tex]v=\sqrt{(20)^2+(14.64)^2}[/tex]
[tex]v=24.79\ m.s^{-1}[/tex]
