What is the kinetic energy of an ideal projectile of mass 19.4 kg at the apex (highest point) of its trajectory, if it was launched with an initial speed of 29.2 m/s and at an initial angle of 46.9° with respect to the horizontal?
I know kinetic energy equals half of mass times velocity squared but the launching at a certain angle is throwing me off here
Vo = 29.2m/s @ 46.9o
Xo = 29.2*Cos46.9 = 20.0 m/s.
Yo = 29.2*sin46.9 = 21.32 m/s.
V = Xo + Yi = 20 + 0i = 20 m/s @ h max.
KE = 0.5m*V^2 = 0.5*19.4*20^2 = 3880 J.
To determine the kinetic energy of the projectile at the apex of its trajectory, we first need to understand the motion of the projectile. Since the projectile is launched at an angle, it has both horizontal and vertical components of motion.
First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component is given by:
Vx = Vi * cos(theta)
where Vx is the horizontal component of velocity, Vi is the initial velocity of the projectile, and theta is the launch angle. Substituting the given values, we have:
Vx = 29.2 m/s * cos(46.9°)
Next, let's determine the vertical component of velocity, Vy. It is given by:
Vy = Vi * sin(theta)
Substituting the given values, we have:
Vy = 29.2 m/s * sin(46.9°)
Now, at the apex of the projectile's trajectory, the vertical component of velocity becomes zero. This means that the projectile reaches its highest point and momentarily stops moving vertically. The horizontal component of velocity remains unchanged throughout the motion.
At the apex, the projectile only has horizontal velocity but no vertical velocity. Therefore, it does not have any kinetic energy at this point because kinetic energy depends on velocity, and the projectile's velocity is zero along the vertical direction.
To summarize, at the apex of its trajectory, the kinetic energy of the projectile is zero.