Answer:
[tex]\vec{v}_{\rm max} = v\cos(\theta)(\^x)\\|\vec{v}_{\rm max}| = v\cos(\theta)\\\vec{a} = \vec{g} = -9.8\^y[/tex]
Explanation:
The equations of kinematics will be used to solve this question:
[tex]y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)\\v_y = v_{y_0} + a_yt[/tex]
At its maximum height, the projectile has zero velocity in the y-direction. But its velocity in the x-direction is unaffected.
First, let's apply the above equations to the x-direction.
There is no acceleration in the x-direction. So, its velocity in the x-direction is constant during the motion.
[tex]v_x = v_{x_0} + a_xt = v_{x_0} + 0\\v_x = v_{x_0} = v\cos(\theta)[/tex]
Therefore, the velocity vector of the projectile is
[tex]v_{max} = v_x = v\cos(\theta)[/tex]
The speed of the projectile is the same.
The acceleration vector is constant during the motion and equal to the gravitational acceleration, which is -9.8 downwards.
