An object is projected with a velocity of 80ms-1 at an angle of 30o to the horizontal.The maximum height reached is

The problem involves finding the maximum height reached by an object that is projected at a given velocity and angle to the horizontal. To solve this, we can use the equations of motion for projectile motion. Firstly, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion and is given by: vx = v * cos(theta) where v is the initial velocity and theta is the angle of projection. The vertical component changes due to the force of gravity and is given by: vy = v * sin(theta) - gt where g is the acceleration due to gravity and t is the time elapsed. At the maximum height, the vertical component of the velocity becomes zero. We can use this fact to determine the time of flight (the time it takes for the object to reach its maximum height and then return to the same height) as follows: 0 = v * sin(theta) - gt_max where t_max is the time taken to reach the maximum height. Rearranging this equation gives: t_max = v * sin(theta) / g Using this value of t_max, we can determine the maximum height reached by the object using the equation: h_max = v * sin(theta) * t_max - 0.5 * g * t_max^2 Substituting the given values in these equations, we get: vx = 80 * cos(30) = 69.28 m/s vy = 80 * sin(30) - 9.81 * t 0 = 80 * sin(30) - 9.81 * t_max t_max = (80 * sin(30)) / 9.81 = 4.08 s h_max = 80 * sin(30) * t_max - 0.5 * 9.81 * t_max^2 = 80 * sin(30) * 4.08 - 0.5 * 9.81 * 4.08^2 = 80 m Therefore, the maximum height reached by the object is 80 meters.