A body is projected with an initial velocity 300m\s at an angle 60 with the vertical;determine
:1.the
velocity at the maximum height
2.The maximum height reached
vertical velocity initial:
300sin60
at top, velocity is only horizontal
vh=300cos60
at top
1/2 m vi'^2=mgh
h= (300sin60)^2/9.8 METERS
To determine the velocity at the maximum height and the maximum height reached, we can use the equations of motion for projectile motion.
1. Velocity at the maximum height:
At the maximum height of a projectile's trajectory, the vertical component of velocity becomes zero. The horizontal component of velocity remains constant throughout the motion.
Given:
Initial velocity (u) = 300 m/s
Angle with the vertical (θ) = 60 degrees
To find the velocity at the maximum height, we need to find the vertical component of the initial velocity.
Vertical component of velocity (v_y) = u * sin(θ)
v_y = 300 m/s * sin(60)
v_y = 300 m/s * (√3/2)
v_y = 300 m/s * 1.732/2
v_y = 259.8 m/s
Therefore, the velocity at the maximum height is 259.8 m/s.
2. Maximum height reached:
To find the maximum height (h), we need to use the equation:
h = (v_y^2) / (2 * g)
where g is the acceleration due to gravity, approximately 9.8 m/s^2.
h = (259.8 m/s)^2 / (2 * 9.8 m/s^2)
h = 67404.04 m^2/s^2 / 19.6 m/s^2
h = 3440.04 m
Therefore, the maximum height reached is approximately 3440.04 meters.