Question

Six baseball throws are shown below. In each case, the ball is thrown with speed at an angle from the horizontal. In all cases, the baseball is thrown from the sameheight above the ground. Assume for the basis of these rankings that the effects of air resistance are negligible.
Rank these throws based on the maximum height reached by the ball.
Rank from largest to smallest.
a. v=15 m/s Theta=60 degrees
b.v=15m/s Theta=45 degrees
c. v=10m/s theta =90 degrees
d. v=10m/s theta=60 degrees
e.v=15 m/s theta =30 degrees
f.v=20 m/s theta =0 degrees

Answer 1

Concepts and reason

The concept required to solve this question is the projectile motion.

Substitute the value of speed, acceleration due to gravity, and the angle in the expression of the time of flight to calculate the time of flight. Follow the same procedure for different cases given in the question. Arrange the time of flights in descending order.

Fundamentals

The expression of the time of flight of the projectile is,

$t = \frac{{2v\sin \theta }}{g}$

Here, t is the time required by the object to hit the ground, v is the speed at which the ball is thrown at an angle $\theta$ from the horizontal, and g is the acceleration due to gravity.

a.

The expression of the time of flight of the projectile is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute 15 m/s for v, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $60^\circ$ for $\theta$.

$\begin{array}{c}\\{t_{\rm{a}}} = \frac{{2\left( {15{\rm{ m/s}}} \right)\sin 60^\circ }}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 2.65{\rm{ s}}\\\end{array}$

The time of flight is 2.65 s.

b.

The expression of the time of flight of the projectile is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute 15 m/s for v, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $45^\circ$ for $\theta$.

$\begin{array}{c}\\{t_{\rm{b}}} = \frac{{2\left( {15{\rm{ m/s}}} \right)\sin 45^\circ }}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 2.16{\rm{ s}}\\\end{array}$

The time of flight is 2.16 s.

c.

The expression of the time of flight of the projectile is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute 10 m/s for v, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $90^\circ$ for $\theta$.

$\begin{array}{c}\\{t_{\rm{c}}} = \frac{{2\left( {10{\rm{ m/s}}} \right)\sin 90^\circ }}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 2.04{\rm{ s}}\\\end{array}$

The time of flight is 2.04 s.

d.

The expression of the time of flight of the projectile is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute 10 m/s for v, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $60^\circ$ for $\theta$.

$\begin{array}{c}\\{t_{\rm{d}}} = \frac{{2\left( {10{\rm{ m/s}}} \right)\sin 60^\circ }}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 1.77{\rm{ s}}\\\end{array}$

The time of flight is 1.77 s.

e.

The expression of the time of flight of the projectile is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute 15 m/s for v, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $30^\circ$ for $\theta$.

$\begin{array}{c}\\{t_{\rm{e}}} = \frac{{2\left( {15{\rm{ m/s}}} \right)\sin 30^\circ }}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 1.53{\rm{ s}}\\\end{array}$

The time of flight is 1.53 s.

f.

The expression of the time of flight of the projectile is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute 20 m/s for v, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $0^\circ$ for $\theta$.

$\begin{array}{c}\\{t_{\rm{f}}} = \frac{{2\left( {{\rm{20 m/s}}} \right)\sin 0^\circ }}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 0{\rm{ s}}\\\end{array}$

The time of flight is 0 s.

g.

Rank the time of flight in the different cases.

The time of flight for the different cases is:

$\begin{array}{l}\\{t_{\rm{a}}} = 2.67{\rm{ s}}\\\\{t_{\rm{b}}} = 2.16{\rm{ s}}\\\\{t_{\rm{c}}} = 2.04{\rm{ s}}\\\\{t_{\rm{d}}} = 1.77{\rm{ s}}\\\end{array}$

And the remaining time of flight is:

$\begin{array}{l}\\{t_{\rm{e}}} = 1.53{\rm{ s}}\\\\{t_{\rm{f}}} = 0{\rm{ s}}\\\end{array}$

Rank from the largest to the smallest is,

${t_{\rm{a}}} > {t_{\rm{b}}} > {t_{\rm{c}}} > {t_{\rm{d}}} > {t_{\rm{e}}} > {t_{\rm{f}}}$

Ans:

Rank of time of flight from the largest to the smallest is: ${t_{\rm{a}}} > {t_{\rm{b}}} > {t_{\rm{c}}} > {t_{\rm{d}}} > {t_{\rm{e}}} > {t_{\rm{f}}}$.