Question

projectile motion

Rank these throws based on the maximum height reached by the ball. Rank from largest to smallest. To rank items as equivalent. overlap them.

Answer 1

General guidance

Concepts and reason

This question is based on the concept of projectile motion.

Projectile motion is the type of motion in which an object is project close to the earth surface and it follows a curved path due the earth\u2019s gravity.

Here, we use the expression of maximum height attained by body in projectile motion.

Substitute different values of u and $\\theta$ in different cases and calculate the maximum height.

In last, compare all the values.

Fundamentals

When an object is thrown at certain angle near the surface of the earth then only gravitational force will act on it (if air resistance is ignored). In other words, in projectile motion vertical component of velocity changes with time but horizontal component of velocity remains constant.

The expression for maximum height of projectile is given as:

$H = \\frac{{{u^2}{{\\sin }^2}\\theta }}{{2g}}$

Here, $g$is the acceleration due to gravity, u is the initial velocity and $\\theta$is the angle of projectile.

Step-by-step

Step 1 of 7

Calculate the maximum height reached by ball.

The expression for maximum height of projectile is,

${H_1} = \\frac{{{u^2}{{\\sin }^2}\\theta }}{{2g}}$

Substitute $15\\,{\\rm{m}} \\cdot {{\\rm{s}}^{ - 1}}$ for $u$and $60^\\circ$ for $\\theta$in above equation.

$\\begin{array}{c}\\\\{H_1} = \\frac{{{{\\left( {15\\,{\\rm{m}} \\cdot {{\\rm{s}}^{ - 1}}} \\right)}^2}{{\\sin }^2}60^\\circ }}{{2\\left( {9.8{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 2}}} \\right)}}\\\\\\\\ = 8.6{\\rm{ m}}\\\\\\end{array}$

Explanation | Common mistakes | Hint for next step

Projectile motion is the type of motion in which an object is project close to the earth surface and it follows a curved path due the earth\u2019s gravity. Maximum height which is the height of the projectile when vertical velocity is zero and the projectile has only its horizontal and vertical components.

Step 2 of 7

Calculate the maximum height$\\left( {{H_2}} \\right)$ reached by ball.

The expression for maximum height of projectile is,

${H_2} = \\frac{{{u^2}{{\\sin }^2}\\theta }}{{2g}}$

Substitute $10{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 1}}$ for $u$and $90^\\circ$ for $\\theta$in above equation.

$\\begin{array}{c}\\\\{H_2} = \\frac{{{{\\left( {10\\,{\\rm{m}} \\cdot {{\\rm{s}}^{ - 1}}} \\right)}^2}{{\\sin }^2}90^\\circ }}{{2\\left( {9.8{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 2}}} \\right)}}\\\\\\\\ = 5.1{\\rm{ m}}\\\\\\end{array}$

Explanation | Common mistakes | Hint for next step

Projectile motion is the type of motion in which an object is project close to the earth surface and it follows a curved path due the earth\u2019s gravity. Maximum height which is the height of the projectile when vertical velocity is zero and the projectile has only its horizontal and vertical components.

Step 3 of 7

Calculate the maximum height$\\left( {{H_3}} \\right)$ reached by ball.

The expression for maximum height of projectile is,

${H_3} = \\frac{{{u^2}{{\\sin }^2}\\theta }}{{2g}}$

Substitute $15{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 1}}$ for $u$and $45^\\circ$ for $\\theta$in above equation.

$\\begin{array}{c}\\\\{H_3} = \\frac{{{{\\left( {15\\,{\\rm{m}} \\cdot {{\\rm{s}}^{ - 1}}} \\right)}^2}{{\\sin }^2}45^\\circ }}{{2\\left( {9.8{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 2}}} \\right)}}\\\\\\\\ = 5.73{\\rm{ m}}\\\\\\end{array}$

Explanation | Common mistakes | Hint for next step

Projectile motion is the type of motion in which an object is project close to the earth surface and it follows a curved path due the earth\u2019s gravity. Maximum height which is the height of the projectile when vertical velocity is zero and the projectile has only its horizontal and vertical components.

Step 4 of 7

Calculate the maximum height$\\left( {{H_4}} \\right)$ reached by ball.

The expression for maximum height of projectile is,

${H_4} = \\frac{{{u^2}{{\\sin }^2}\\theta }}{{2g}}$

Substitute $10{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 1}}$ for $u$and $60^\\circ$ for $\\theta$in above equation.

$\\begin{array}{c}\\\\{H_4} = \\frac{{{{\\left( {10\\,{\\rm{m}} \\cdot {{\\rm{s}}^{ - 1}}} \\right)}^2}{{\\sin }^2}60^\\circ }}{{2\\left( {9.8{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 2}}} \\right)}}\\\\\\\\ = 3.82{\\rm{ m}}\\\\\\end{array}$

Explanation | Common mistakes | Hint for next step

Projectile motion is the type of motion in which an object is project close to the earth surface and it follows a curved path due the earth\u2019s gravity. Maximum height which is the height of the projectile when vertical velocity is zero and the projectile has only its horizontal and vertical components.

Step 5 of 7

Calculate the maximum height$\\left( {{H_5}} \\right)$ reached by ball.

The expression for maximum height of projectile is,

${H_5} = \\frac{{{u^2}{{\\sin }^2}\\theta }}{{2g}}$

Substitute $15{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 1}}$ for $u$and $30^\\circ$ for $\\theta$in above equation.

$\\begin{array}{c}\\\\{H_5} = \\frac{{{{\\left( {15\\,{\\rm{m}} \\cdot {{\\rm{s}}^{ - 1}}} \\right)}^2}{{\\sin }^2}30^\\circ }}{{2\\left( {9.8{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 2}}} \\right)}}\\\\\\\\ = 2.86{\\rm{ m}}\\\\\\end{array}$

Explanation | Common mistakes | Hint for next step

Projectile motion is the type of motion in which an object is project close to the earth surface and it follows a curved path due the earth\u2019s gravity. Maximum height which is the height of the projectile when vertical velocity is zero and the projectile has only its horizontal and vertical components.

Step 6 of 7

Calculate the maximum height$\\left( {{H_6}} \\right)$ reached by ball.

The expression for maximum height of projectile is,

${H_6} = \\frac{{{u^2}{{\\sin }^2}\\theta }}{{2g}}$

Substitute $20{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 1}}$ for $u$and $0^\\circ$ for $\\theta$in above equation.

$\\begin{array}{c}\\\\{H_6} = \\frac{{{{\\left( {20\\,{\\rm{m}} \\cdot {{\\rm{s}}^{ - 1}}} \\right)}^2}{{\\sin }^2}0^\\circ }}{{2\\left( {9.8{\\rm{ m}} \\cdot {{\\rm{s}}^{ - 2}}} \\right)}}\\\\\\\\ = 0{\\rm{ m}}\\\\\\end{array}$

Explanation | Common mistakes | Hint for next step

Projectile motion is the type of motion in which an object is project close to the earth surface and it follows a curved path due the earth\u2019s gravity. Maximum height which is the height of the projectile when vertical velocity is zero and the projectile has only its horizontal and vertical components.

Step 7 of 7

Rank from largest to smallest maximum height reached by the ball.

Maximum height reached by the ball $\\left( {{H_1}} \\right)$is $8.6{\\rm{ m}}$

Maximum height reached by the ball $\\left( {{H_2}} \\right)$is $5.1{\\rm{ m}}$

Maximum height reached by the ball $\\left( {{H_3}} \\right)$is$5.73{\\rm{ m}}$

Maximum height reached by the ball $\\left( {{H_4}} \\right)$is$3.82{\\rm{ m}}$

Maximum height reached by the ball $\\left( {{H_5}} \\right)$is$2.86{\\rm{ m}}$

Maximum height reached by the ball $\\left( {{H_6}} \\right)$is $0{\\rm{ m}}$

Rank from largest to smallest.

$\\left( {{H_1}} \\right)$>$\\left( {{H_3}} \\right)$>$\\left( {{H_2}} \\right)$>$\\left( {{H_4}} \\right)$>$\\left( {{H_5}} \\right)$>$\\left( {{H_6}} \\right)$

Rank from largest to smallest is $\\left( {{H_1}} \\right)$>$\\left( {{H_3}} \\right)$>$\\left( {{H_2}} \\right)$>$\\left( {{H_4}} \\right)$>$\\left( {{H_5}} \\right)$>$\\left( {{H_6}} \\right)$.

Explanation

Projectile motion is the type of motion in which an object is project close to the earth surface and it follows a curved path due the earth\u2019s gravity. Ranking based on the values of above obtained heights from largest to smallest.

Answer

Rank from largest to smallest is $\\left( {{H_1}} \\right)$>$\\left( {{H_3}} \\right)$>$\\left( {{H_2}} \\right)$>$\\left( {{H_4}} \\right)$>$\\left( {{H_5}} \\right)$>$\\left( {{H_6}} \\right)$.

Answer only

Rank from largest to smallest is $\\left( {{H_1}} \\right)$>$\\left( {{H_3}} \\right)$>$\\left( {{H_2}} \\right)$>$\\left( {{H_4}} \\right)$>$\\left( {{H_5}} \\right)$>$\\left( {{H_6}} \\right)$.