Respuesta :
Answer:
The maximum height reached by the toy rocket is 0.73 meters.
Explanation:
It is given that,
Mass of the toy rocket, m = 0.8 kg
It is projected at an angle of 45 degrees from ground level. The kinetic energy of the rocket is used to find the velocity with which it was projected as :
[tex]K=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 41}{0.8}}[/tex]
u = 10.12 m/s
When it reaches to a maximum height, its final velocity will be equal to 0, v = 0
Using third equation of motion to find it as :
[tex]v^2-u^2=2as[/tex]
a = -g
[tex]-u^2=-2gs[/tex]
s is the maximum height reached by the toy rocket
In vertical direction, [tex]u_y=u\ sin\theta[/tex]
[tex]s=\dfrac{u_y}{2g}[/tex]
[tex]s=\dfrac{u\ sin\theta}{2g}[/tex]
[tex]s=\dfrac{10.12\times sin(45)}{2\times 9.8}[/tex]
s = 0.36 meters
So, the maximum height reached by the toy rocket is 0.73 meters. Hence, this is the required solution.
