Example Question - time of flight
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Calculating Time to Reach Zero Speed and Return Time for a Ball Thrown Upward
The first part of the problem asks for the time it will take for the ball to reach zero speed when thrown upward at a speed of \(4.5 \ ms^{-1}\). The motion of the ball can be described by the kinematic equation which includes the initial velocity (\(v_i\)), the final velocity (\(v_f\)), the acceleration due to gravity (\(g\)), and the time (\(t\)): \[v_f = v_i + at\] For the ball to come to a stop at the highest point of its trajectory, its final velocity, \(v_f\), would be \(0 \ ms^{-1}\), and the acceleration, \(a\), would be \(-9.8 \ ms^{-2}\) as it's going against gravity. By substituting these values and the given initial velocity \(v_i = 4.5 \ ms^{-1}\) into the equation above: <p>\(0 = 4.5 \ ms^{-1} - (9.8 \ ms^{-2}) \times t \)</p> <p>\(9.8 \ ms^{-2} \times t = 4.5 \ ms^{-1} \)</p> <p>\(t = \frac{4.5 \ ms^{-1}}{9.8 \ ms^{-2}} \)</p> <p>\(t \approx 0.459 \ s\)</p> This is the time it takes for the ball to reach the highest point where its speed is zero. The second part of the problem asks for the time it will take for the ball to return to its starting point. This can be calculated by considering the symmetry of projectile motion. The time taken to ascend to the highest point is the same as the time taken to descent back to the starting point. Therefore, the total time for the whole journey is double the time calculated for the upward journey: <p>\(Total\ time = 2 \times \text{time to reach maximum height} \)</p> <p>\(Total\ time = 2 \times 0.459 \ s \)</p> <p>\(Total\ time \approx 0.918 \ s \)</p> The third part of the problem asks for the final speed of the ball when it returns to its starting point. The speed of the ball when it returns to the starting point will be the same as the speed it had when it was thrown upward, but in the opposite direction (assuming no air resistance). Therefore: <p>\(Final\ speed = 4.5 \ ms^{-1}\)</p>
Projectile Motion Calculations
// Problema 1 a) Magnitud de la velocidad a los 4 segundos: <p>$$ v = v_0 + g \cdot t = 6 \, m/s + (9.8 \, m/s^2)(4 \, s) = 45.2 \, m/s $$</p> b) Distancia recorrida entre los segundos 4 y 5: <p>$$ d = v_0 \cdot t + \frac{1}{2} g \cdot t^2 $$</p> <p>$$ d_{4s} = (6 \, m/s)(4 \, s) + \frac{1}{2}(9.8 \, m/s^2)(4 \, s)^2 = 24 \, m + 78.4 \, m = 102.4 \, m $$</p> <p>$$ d_{5s} = (6 \, m/s)(5 \, s) + \frac{1}{2}(9.8 \, m/s^2)(5 \, s)^2 = 30 \, m + 122.5 \, m = 152.5 \, m $$</p> <p>$$ d_{4\_5s} = d_{5s} - d_{4s} = 152.5 \, m - 102.4 \, m = 50.1 \, m $$</p> // Problema 2 a) Distancia recorrida a los 3 segundos: <p>$$ d = v_0 \cdot t - \frac{1}{2} g \cdot t^2 = 30 \, m/s \cdot 3 \, s - \frac{1}{2}(9.8 \, m/s^2)(3 \, s)^2 = 90 \, m - 44.1 \, m = 45.9 \, m $$</p> b) Magnitud de la velocidad a los 3 segundos: <p>$$ v = v_0 - g \cdot t = 30 \, m/s - (9.8 \, m/s^2)(3 \, s) = 30 \, m/s - 29.4 \, m/s = 0.6 \, m/s $$</p> c) Altura máxima alcanzada (cuando \( v = 0 \)): <p>$$ 0 = v_0^2 - 2 g \cdot d_{max} $$</p> <p>$$ d_{max} = \frac{v_0^2}{2g} = \frac{(30 \, m/s)^2}{2(9.8 \, m/s^2)} = \frac{900}{19.6} \approx 45.9 \, m $$</p> d) El tiempo que tardará en el aire (tiempo hasta subir y bajar): <p>$$ t_{subida} = \frac{v_0}{g} = \frac{30 \, m/s}{9.8 \, m/s^2} \approx 3.06 \, s $$</p> <p>$$ t_{total} = 2 \cdot t_{subida} \approx 2 \cdot 3.06 \, s = 6.12 \, s $$</p>
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