Question - Calculating Maximum Height of a Ball Thrown Upwards
Solution:
The image contains a handwritten question which states:"A ball is thrown from the ground to the top of a school flagpole. If it returns to the ground after 4s. What is the height?"To solve this problem, we can use the kinematic equation for the position (height in this case) as a function of time while an object is in free fall under the influence of gravity. The equation is:\[ h = v_i t + \frac{1}{2} a t^2 \]where:- $$ h $$ is the height,- $$ v_i $$ is the initial velocity,- $$ t $$ is the time,- $$ a $$ is the acceleration due to gravity (which is approximately 9.8 m/s² downwards).Given that the round trip takes 4 seconds, the ball takes 2 seconds to reach the maximum height (since the time going up is the same as the time going down for a symmetric trajectory under gravity).Assuming that the ball is thrown and caught at the same height (ground level) and neglecting air resistance, we can take the acceleration due to gravity as -9.8 m/s² (negative because it's acting downwards).Let's set the time for half the trip (just to reach the peak) as:\[ t = 2 \text{ seconds} \]And since the ball stops at the peak (the top of its trajectory), the initial velocity can be found by using the equation (at the peak the final velocity is 0 m/s):\[ v_f = v_i + at \]\[ 0 = v_i - 9.8 \times 2 \]\[ v_i = 19.6 \text{ m/s} \]Now we can calculate the height $$ h $$ using the initial velocity $$ v_i $$ and half the total time (up trip only):\[ h = v_i t + \frac{1}{2} a t^2 \]\[ h = 19.6 \times 2 + \frac{1}{2} (-9.8) \times 2^2 \]\[ h = 39.2 - 19.6 \]\[ h = 19.6 \text{ meters} \]Therefore, the height of the school flagpole, or the maximum height achieved by the ball, is 19.6 meters.
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