In a physics experiment, a ball is thrown upwards with an initial velocity of 49 m/s. Ignoring air resistance, how long will it take for the ball to reach its highest point? (Use \(g = 9.8 \, \text{m/s}^2\) for acceleration due to gravity.)

["How Long Does a Ball Take to Reach Its Highest Point When Thrown Upward?", "Understanding the motion of objects thrown into the air is fundamental in physics, and one classic experiment involves launching a ball straight upward with an initial velocity. If a ball is thrown upward with an initial velocity of 49 m/s and we ignore air resistance, we can precisely calculate how long it takes to reach its peak height. This event marks the moment when the ball momentarily stops before descending.", "In projectile motion, the time to reach the highest point occurs precisely when the object’s vertical velocity becomes zero—before gravity pulls it back down. Ignoring air resistance simplifies the analysis, allowing us to apply basic kinematic equations.", "Key Physics Concept: Deceleration Due to Gravity", "The ball decelerates uniformly under gravity’s constant downward acceleration, ( g = 9.8 , \ ext{m/s}^2 ). At the highest point, the vertical velocity is zero. We use the equation:", "[\nv = u - gt\n]", "where:\n- ( v ) = final velocity (0 m/s at peak),\n- ( u ) = initial upward velocity (49 m/s),\n- ( g ) = acceleration due to gravity (9.8 m/s²),\n- ( t ) = time to reach peak.", "Rearranging to solve for ( t ):", "[\nt = \frac{v - u}{-g} = \frac{0 - 49}{-9.8} = \frac{49}{9.8} = 5 , \ ext{seconds}\n]", "Conclusion", "It takes exactly 5 seconds for the ball to reach its highest point when thrown upward at 49 m/s in the absence of air resistance. At this moment, the ball’s vertical velocity is zero, and it begins its descent under gravitational pull. This straightforward calculation underscores the power of kinematics in predicting motion, making it a key concept in physics experiments and real-world applications involving vertical motion."]