Respuesta :
Answer:
a) H = 26.63 m.
b) [tex]y_{max} = 34.13~m[/tex]
c) v = 18.89 m/s.
Explanation:
a) We will use kinematics equations for 2D-projectile motion.
First we will separate the velocity into x- and y-components, and apply the kinematics equations separately.
y-direction:
[tex]y - y_0 = v_{y_0}t + \frac{1}{2}at^2\\H - 1.95 = (29\sin(60))(3.8) + \frac{1}{2}(-9.8)(3.8)^2\\H = 26.63~m[/tex]
b) Another kinematics equation will help us to find the maximum height.
[tex]v_y^2 = v_{y_0}^2 + 2a(\Delta y)\\0 = (29\sin(60))^2 + 2(-9.8)(y_{max} - 1.95)\\y_{max} = 34.13~m[/tex]
c) The impact speed is the sum of the speed in both directions.
[tex]v_y^2 = v_{y_0}^2 + 2a(H-y_0)\\v_y^2 = (29\sin(60))^2 + 2(-9.8)(26.63 - 1.95)\\v_y = 12.12~m/s[/tex]
[tex]v_x^2 = v_{x_0}^2 + 2a_x(\Delta x)\\v_x^2 = (29\cos(60))^2 + 2(0)(\Delta x)\\v_x = 14.5~m/s[/tex]
The impact speed is
[tex]v = \sqrt{v_x^2 + v_y^2} = \sqrt{14.5^2 + 12.12^2} = 18.89~m/s[/tex]
