Respuesta :
Answer:
The maximum height reached by both balls are 9685.93 meters and 32653.06 meters respectively.
Explanation:
Given that,
Mass of the first cannonball, [tex]m_1=19\ kg[/tex]
Initial speed of the first cannonball, [tex]u_1=800\ m/s[/tex]
Angle of projection of the first cannonball, [tex]\theta_1=33^{\circ}[/tex]
Angle of projection of the second cannonball, [tex]\theta_2=90^{\circ}[/tex]
In this case, we need to find the maximum height reached by each ball. We know that the maximum height reached by a projectile is given by :
[tex]h_1=\dfrac{(u\ \sin\theta_1)^2}{2g}\\\\h_1=\dfrac{(800\ \sin(33))^2}{2\times 9.8}\\\\h_1=9685.93\ m[/tex]
The maximum height reached by the second ball is given by :
[tex]h_2=\dfrac{(u\ \sin\theta_2)^2}{2g}\\\\h_2=\dfrac{(800\ \sin(90))^2}{2\times 9.8}\\\\h_2=32653.06\ m[/tex]
So, the maximum height reached by both balls are 9685.93 meters and 32653.06 meters respectively. Hence, this is the required solution.
