Respuesta :
Answer:
The maximum height covered is 3.25 m.
The horizontal distance covered is 9.81 m.
The total time in the air is 1.63 seconds.
Explanation:
The launch speed, [tex]u_0= 10 m/s[/tex].
Angle of launch with the horizontal, [tex]\theta = 53 ^{\circ}[/tex]
So, the vertical component of the initial velocity,
[tex]u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)[/tex].
The horizontal component of the initial velocity,
[tex]u_0\cos\theta=10 \cos 53 ^{\circ}[/tex]
Let, t be the time of flight, to the horizontal distance covered
[tex]D=10 \cos (53 ^{\circ})t\cdots(ii)[/tex].
Not, applying the equation of motion in the vertical direction.
[tex]s= ut +\frac 1 2 at^2[/tex]
Where s is the displacement in time t, u is the initial velocity and a is the acceleration.
In this case, [tex]u =10 \sin 53 ^{\circ}[/tex] (from equation (i), s=0 (as the final height is same as the launch height) and [tex]a = -9.81 m/s^2[/tex] (negative sign is due to the downward direction).
[tex]\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2[/tex]
[tex]\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63[/tex] seconds.
So, the total time in the air is 1.63 seconds.
From equation (i),
Total horizontal distance covered is
[tex]D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m[/tex].
Now, for the maximum height, H, applying the equation of motion as
[tex]v^2=u^2+2as[/tex]
Here, v is the final velocity and v=0 (at the maximum height), and h=H.
So, [tex]0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H[/tex]
[tex]\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}[/tex]
[tex]\Rightarrow H = 3.25 m[/tex].
Hence, the maximum height covered is 3.25 m.
