Ig note the effects of air resistance. 1. A projectile is fired vertically upward from a motionless balloon in the air. The projectile leaves the balloon at a velocity of \( 200 \mathrm{~m} \mathrm{~s}^{-1} \) and strikes the ground at ci -300 ms Calculates a) the height of the balloonabove the ground b) the time it takes for the projectile to reach the ground, 2. A metal sphere is dropped fromaballoon, which is ning at a constant velocity of 5 msi The metal spherestrikes the ground after 5 seconds. Calculate: a) the velocity with which the sphere strikes the ground b) how far above the growind the balloon was When the sphere was released 3. Ahot ari balloon moves vertically upward at a constant veloaty of \( \overline{6} \mathrm{~m}^{-1} \). When the baloon is 87 mabove the ground a botte is thrown upward from inside the balloonat 3 ms ? (a) What is the max amim heightreached by the botte? (b) Catculate the time taken for the botte to reach the ground

Let's break down the problems step by step. ### Problem 1: Projectile Fired Vertically Upward 1. **Known Conditions:** - Initial velocity of the projectile, \( u = 200 \, \text{m/s} \) - Final velocity when it strikes the ground, \( v = -300 \, \text{m/s} \) (downward) - Acceleration due to gravity, \( g = -9.81 \, \text{m/s}^2 \) (acting downward) #### a) Calculate the height of the balloon above the ground Using the kinematic equation: \[ v^2 = u^2 + 2as \] where: - \( v \) = final velocity - \( u \) = initial velocity - \( a \) = acceleration - \( s \) = displacement (height of the balloon) Rearranging for \( s \): \[ s = \frac{v^2 - u^2}{2a} \] Substituting the values: \[ s = \frac{(-300)^2 - (200)^2}{2 \times (-9.81)} \] #### b) Calculate the time it takes for the projectile to reach the ground Using the kinematic equation: \[ v = u + at \] Rearranging for \( t \): \[ t = \frac{v - u}{a} \] Substituting the values: \[ t = \frac{-300 - 200}{-9.81} \] ### Problem 2: Metal Sphere Dropped from a Balloon 1. **Known Conditions:** - Initial velocity of the sphere, \( u = 5 \, \text{m/s} \) (downward) - Time of fall, \( t = 5 \, \text{s} \) - Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \) #### a) Calculate the velocity with which the sphere strikes the ground Using the kinematic equation: \[ v = u + gt \] Substituting the values: \[ v = 5 + 9.81 \times 5 \] #### b) Calculate how far above the ground the balloon was when the sphere was released Using the equation for displacement: \[ s = ut + \frac{1}{2}gt^2 \] Substituting the values: \[ s = 5 \times 5 + \frac{1}{2} \times 9.81 \times (5^2) \] ### Problem 3: Hot Air Balloon and Bottle Thrown Upward 1. **Known Conditions:** - Velocity of the balloon, \( v_b = 6 \, \text{m/s} \) (upward) - Height of the balloon, \( h = 87 \, \text{m} \) - Initial velocity of the bottle, \( u = 3 \, \text{m/s} \) (upward) - Acceleration due to gravity, \( g = -9.81 \, \text{m/s}^2 \) #### a) Calculate the maximum height reached by the bottle Using the kinematic equation: \[ v^2 = u^2 + 2as \] At maximum height, \( v = 0 \): \[ 0 = (3)^2 + 2(-9.81)s \] Rearranging for \( s \): \[ s = \frac{(3)^2}{2 \times 9.81} \] The total height above the ground will be: \[ \text{Total height} = h + s \] #### b) Calculate the time taken for the bottle to reach the ground Using the equation: \[ s = ut + \frac{1}{2}gt^2 \] where \( s = -h \) (the bottle falls back to the ground): \[ -h = ut + \frac{1}{2}gt^2 \] Substituting the values: \[ -87 = 3t - \frac{1}{2} \times 9.81t^2 \] Now, let's perform the calculations for each part. Calculate the value by following steps: - step0: Calculate: \(5+9.81\times 5\) - step1: Multiply the numbers: \(5+49.05\) - step2: Add the numbers: \(54.05\) Calculate or simplify the expression \( -87 = 3t - (1/2) * 9.81 * t^2 \). Calculate the value by following steps: - step0: Calculate: \(5\times 5+\frac{1}{2}\times 9.81\times 5^{2}\) - step1: Multiply the numbers: \(25+\frac{1}{2}\times 9.81\times 5^{2}\) - step2: Multiply: \(25+\frac{981}{8}\) - step3: Reduce fractions to a common denominator: \(\frac{25\times 8}{8}+\frac{981}{8}\) - step4: Transform the expression: \(\frac{25\times 8+981}{8}\) - step5: Multiply the numbers: \(\frac{200+981}{8}\) - step6: Add the numbers: \(\frac{1181}{8}\) Calculate or simplify the expression \( (3^2) / (2 * 9.81) \). Calculate the value by following steps: - step0: Calculate: \(\frac{3^{2}}{\left(2\times 9.81\right)}\) - step1: Remove the parentheses: \(\frac{3^{2}}{2\times 9.81}\) - step2: Multiply the numbers: \(\frac{3^{2}}{19.62}\) - step3: Convert the expressions: \(\frac{3^{2}}{\frac{981}{50}}\) - step4: Evaluate the power: \(\frac{9}{\frac{981}{50}}\) - step5: Multiply by the reciprocal: \(9\times \frac{50}{981}\) - step6: Reduce the numbers: \(1\times \frac{50}{109}\) - step7: Multiply: \(\frac{50}{109}\) Calculate or simplify the expression \( ((-300)^2 - (200)^2) / (2 * -9.81) \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(\left(-300\right)^{2}-200^{2}\right)}{\left(2\left(-9.81\right)\right)}\) - step1: Remove the parentheses: \(\frac{\left(-300\right)^{2}-200^{2}}{2\left(-9.81\right)}\) - step2: Multiply the numbers: \(\frac{\left(-300\right)^{2}-200^{2}}{-19.62}\) - step3: Evaluate the power: \(\frac{300^{2}-40000}{-19.62}\) - step4: Convert the expressions: \(\frac{300^{2}-40000}{-\frac{981}{50}}\) - step5: Multiply by the reciprocal: \(\left(300^{2}-40000\right)\left(-\frac{50}{981}\right)\) - step6: Multiply the numbers: \(-\left(300^{2}-40000\right)\times \frac{50}{981}\) - step7: Multiply: \(-\frac{\left(300^{2}-40000\right)\times 50}{981}\) - step8: Multiply: \(-\frac{2500000}{981}\) Calculate or simplify the expression \( (-300 - 200) / -9.81 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(-300-200\right)}{\left(-9.81\right)}\) - step1: Remove the parentheses: \(\frac{-300-200}{-9.81}\) - step2: Subtract the numbers: \(\frac{-500}{-9.81}\) - step3: Convert the expressions: \(\frac{-500}{-\frac{981}{100}}\) - step4: Multiply by the reciprocal: \(-500\left(-\frac{100}{981}\right)\) - step5: Use the rules for multiplication and division: \(500\times \frac{100}{981}\) - step6: Multiply: \(\frac{500\times 100}{981}\) - step7: Multiply: \(\frac{50000}{981}\) Solve the equation \( -87 = 3t - (1/2) * 9.81 * t^2 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(-87=3t-\frac{1}{2}\times 9.81t^{2}\) - step1: Multiply the terms: \(-87=3t-\frac{981}{200}t^{2}\) - step2: Swap the sides: \(3t-\frac{981}{200}t^{2}=-87\) - step3: Move the expression to the left side: \(3t-\frac{981}{200}t^{2}+87=0\) - step4: Rewrite in standard form: \(-\frac{981}{200}t^{2}+3t+87=0\) - step5: Multiply both sides: \(\frac{981}{200}t^{2}-3t-87=0\) - step6: Multiply both sides: \(200\left(\frac{981}{200}t^{2}-3t-87\right)=200\times 0\) - step7: Calculate: \(981t^{2}-600t-17400=0\) - step8: Solve using the quadratic formula: \(t=\frac{600\pm \sqrt{\left(-600\right)^{2}-4\times 981\left(-17400\right)}}{2\times 981}\) - step9: Simplify the expression: \(t=\frac{600\pm \sqrt{\left(-600\right)^{2}-4\times 981\left(-17400\right)}}{1962}\) - step10: Simplify the expression: \(t=\frac{600\pm \sqrt{600^{2}+68277600}}{1962}\) - step11: Simplify the expression: \(t=\frac{600\pm 60\sqrt{19066}}{1962}\) - step12: Separate into possible cases: \(\begin{align}&t=\frac{600+60\sqrt{19066}}{1962}\\&t=\frac{600-60\sqrt{19066}}{1962}\end{align}\) - step13: Simplify the expression: \(\begin{align}&t=\frac{100+10\sqrt{19066}}{327}\\&t=\frac{600-60\sqrt{19066}}{1962}\end{align}\) - step14: Simplify the expression: \(\begin{align}&t=\frac{100+10\sqrt{19066}}{327}\\&t=\frac{100-10\sqrt{19066}}{327}\end{align}\) - step15: Rewrite: \(t_{1}=\frac{100-10\sqrt{19066}}{327},t_{2}=\frac{100+10\sqrt{19066}}{327}\) Let's summarize the results of the calculations for each part of the problems. ### Problem 1: Projectile Fired Vertically Upward #### a) Height of the balloon above the ground Using the kinematic equation: \[ s = \frac{(-300)^2 - (200)^2}{2 \times (-9.81)} \approx 2548.42 \, \text{m} \] #### b) Time it takes for the projectile to reach the ground Using the kinematic equation: \[ t = \frac{-300 - 200}{-9.81} \approx 50.97 \, \text{s} \] ### Problem 2: Metal Sphere Dropped from a Balloon #### a) Velocity with which the sphere strikes the ground Using the kinematic equation: \[ v = 5 + 9.81 \times 5 \approx 54.05 \, \text{m/s} \] #### b) How far above the ground the balloon was when the sphere was released Using the displacement equation: \[ s = 5 \times 5 + \frac{1}{2} \times 9.81 \times (5^2) \approx 147.625 \, \text{m} \] ### Problem 3: Hot Air Balloon and Bottle Thrown Upward #### a) Maximum height reached by the bottle Using the kinematic equation: \[ s = \frac{(3)^2}{2 \times 9.81} \approx 0.4587 \, \text{m} \] The total height above the ground will be: \[ \text{Total height} = 87 + 0.4587 \approx 87.4587 \, \text{m} \] #### b) Time taken for the bottle to reach the ground The equation to solve is: \[ -87 = 3t - \frac{1}{2} \times 9.81 \times t^2 \] The solutions for \( t \) are: \[ t_{1} = \frac{100 - 10\sqrt{19066}}{327}, \quad t_{2} = \frac{100 + 10\sqrt{19066}}{327} \] Calculating these values will give the time taken for the bottle to reach the ground. ### Summary of Results 1. **Projectile Fired Vertically Upward:** - Height of the balloon: \( \approx 2548.42 \, \text{m} \) - Time to reach the ground: \( \approx 50.97 \, \text{s} \) 2. **Metal Sphere Dropped from a Balloon:** - Velocity when striking the ground: \( \approx 54.05 \, \text{m/s} \) - Height of the balloon: \( \approx 147.625 \, \text{m} \) 3. **Hot Air Balloon and Bottle:** - Maximum height reached by the bottle: \( \approx 87.4587 \, \text{m} \) - Time taken for the bottle to reach the ground: \( t_{1} \) and \( t_{2} \) (exact values depend on further calculation). If you need further calculations for the time taken for the bottle to reach the ground, please let me know!