Example Question - maximum height
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Determining the Time and Maximum Height of a Projectiles Launch
<p>To determine the time at which the projectile reaches its maximum height, we need to find when the velocity is zero. The velocity is the derivative of the position function \( y(t) \).</p> <p>The position function given in the question is:</p> <p>\[ y(t) = -16t^2 + 64t + 80 \]</p> <p>To find the velocity, we differentiate \( y(t) \) with respect to \( t \):</p> <p>\[ v(t) = y'(t) = \frac{d}{dt}(-16t^2 + 64t + 80) = -32t + 64 \]</p> <p>We set the velocity to zero and solve for \( t \):</p> <p>\[ 0 = -32t + 64 \]</p> <p>\[ 32t = 64 \]</p> <p>\[ t = 2 \]</p> <p>The projectile reaches its maximum height at \( t = 2 \) seconds.</p> <p>Now we calculate the maximum height by substituting \( t = 2 \) back into the position function \( y(t) \):</p> <p>\[ y(2) = -16(2)^2 + 64(2) + 80 \]</p> <p>\[ y(2) = -16(4) + 128 + 80 \]</p> <p>\[ y(2) = -64 + 128 + 80 \]</p> <p>\[ y(2) = 144 \]</p> <p>Therefore, the maximum height reached by the projectile is 144 feet.</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>
Calculating the Maximum Height Reached by a Ball Thrown Vertically Upward
<p>To find the maximum height \( h \) reached by the ball, we use the kinematics equation which relates the final velocity \( v \), initial velocity \( u \), acceleration \( a \), and displacement \( s \) (in this case, height \( h \)):</p> <p>\[ v^2 = u^2 + 2as \]</p> <p>At the maximum height, the final velocity \( v \) will be 0 m/s (since the ball stops rising before it starts to fall). The initial velocity \( u \) is 10 m/s (upward), and the acceleration \( a \) due to gravity is -10 m/s² (downward). Thus, we can plug these into our equation:</p> <p>\[ 0 = (10)^2 + 2(-10)h \]</p> <p>Which simplifies to:</p> <p>\[ 0 = 100 - 20h \]</p> <p>Now we solve for \( h \):</p> <p>\[ 20h = 100 \]</p> <p>\[ h = \frac{100}{20} \]</p> <p>\[ h = 5 \]</p> <p>Thus, the maximum height \( h \) reached by the ball is 5 meters.</p>
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