A Ball Is Thrown Directly Upwards At A Velocity Of 20 M/S

A Ball Is Thrown Directly Upwards At A Velocity Of 20ms From A Hei

Calculate the total time it takes for the ball to reach the ground, given it is thrown vertically upwards at 20 m/s from a height of 2 meters, assuming no air resistance.

The problem involves analyzing the projectile's motion under uniform acceleration due to gravity. The acceleration due to gravity (g) is approximately 9.81 m/s² downward. The initial velocity (u) of the ball is 20 m/s upwards, and the initial height (h₀) is 2 meters.

First, determine the time taken for the upward motion until the ball reaches its peak: at the peak, vertical velocity becomes zero. Using v = u - g t, where v = 0 at the maximum height, we find t_up = u / g = 20 / 9.81 ≈ 2.04 seconds.

Next, compute the maximum height reached above the initial point: h_max = h₀ + u t_up - (1/2) g t_up². Substituting yields h_max ≈ 2 + 20 2.04 - 0.5 9.81 * (2.04)² ≈ 2 + 40.8 - 20.4 ≈ 22.4 meters above the initial position.

Now, calculate the total time to fall from this maximum height back to the ground. The overall height from the maximum point down to the ground level is h_total = h_max + initial height offset, but since initial height is included, the total from the maximum height to the ground is h_max + h₀ = 22.4 + 2 = 24.4 meters.

Using the kinematic equation for free fall: 0 = v_0² - 2 g h, where v_0 = 0 at impact, rearranged for time as t_down = sqrt(2 h / g). Plugging in h = 22.4 meters, approximates to t_down ≈ sqrt(2 * 22.4 / 9.81) ≈ 2.14 seconds.

The total time for the complete trajectory is T_total = t_up + t_down ≈ 2.04 + 2.14 ≈ 4.18 seconds.

Paper For Above instruction

The problem involves timing the projectile's entire journey from launch to landing, taking into account the upward motion, the peak, and the downward travel. First, initial parameters are used to find the ascent duration by setting vertical velocity to zero. Due to symmetry in projectile motion, the time to reach maximum height can be calculated as u / g, leading to approximately 2.04 seconds. The maximum height above the initial point is derived from kinematic equations, resulting in about 22.4 meters above the starting position. From this point, the object falls to the ground, covering a distance of approximately 24.4 meters. Using the fall equation, the descent time approximates at 2.14 seconds. Summing both ascent and descent times, the total flight duration is approximately 4.18 seconds.

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