Chapter 3, Problem 16Q

Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with larger than 6B. (a) Which cannonball reaches a higher elevation? (b) Which stays longer in the air? (c) Which travels farther? Explain.

Solution

To Determine: Two cannon balls, A and B are fired from the ground with identical initial speed, but with ${\theta }_A\text{ larger than }{\theta }_B$ .

(a) Thecannonballwhich reaches a higher elevation.

(b) The one that longer in the air.

(c) The one that travels farther, along with anexplanation.

Solution:

(a)Cannonball A reaches a higher elevation.

(b)CannonballA stays longer in the air.

(c)We can’t say anything which travels farther to compare we should have a numerical value of ${\theta }_A\text{ and }{\theta }_B$ .

Explanation:

Given:

The initial speed of both cannonballs A and B are same.

  $v_A=v_B=v_i$

Angle of projection of cannon ball A $={\theta }_A$

Angle of projection of cannon ball A $={\theta }_B$

Formula Used:

Horizontal motion is uniform motion. Vertical motion is motion with constant velocity.

Time of flight, $T_f=\frac{2v_i\sin \theta }{g}\mathrm{...}\left(1\right)$

Maximum height reached by projectile, $h_{\max }=\frac{v_i{}^2{\sin }^2\theta }{2g}\mathrm{...}(2)$

Range of projectile, $\text{R=}\frac{v_i{}^2\sin 2\theta }{g}\mathrm{...}(3)$

Explanation:

(a)

  $h_{\max }=\frac{v_i{}^2{\sin }^2\theta }{2g}$

  $h_{\max 1}=\frac{v_i{}^2{\sin }^2{\theta }_A}{2g}$ , $h_{\max 2}=\frac{v_i{}^2{\sin }^2{\theta }_B}{2g}$

  ${\theta }_A>{\theta }_B$

  $\begin{array}{l}\sin {\theta }_A>\sin {\theta }_B\\ {\sin }^2{\theta }_A>{\sin }^2{\theta }_B\\ \frac{v_i{}^2{\sin }^2{\theta }_A}{2g}>\frac{v_i{}^2{\sin }^2{\theta }_B}{2g}\\ h_{\max ,1}>h_{\max ,2}\end{array}$

Hence, Cannonballs A reaches a higher elevation.

(b)

Time of flight, $T_f=\frac{2v_i\sin \theta }{g}$

Time of flight of cannonball A, $\left(T_A\right)=\frac{2v_i\sin {\theta }_A}{g}$

Time of flight of cannonball B, $\left(T_B\right)=\frac{2v_i\sin {\theta }_B}{g}$

  $T_A>T_B$

Cannonball A stays longer in the air.

(c)

  $\begin{array}{l}\text{Range}\left(R\right)=\frac{v_i{}^2\sin 2\theta }{g}\\ R_A=\frac{v_i{}^2\sin 2{\theta }_A}{g},{\text{ R}}_B=\frac{v_i{}^2\sin 2{\theta }_B}{g}\end{array}$

Case1:

If ${45}^o\ge {\theta }_A>{\theta }_B$

Then, $\sin 2{\theta }_A>\sin 2\theta$

So,

  $\frac{v_i{}^2\sin 2{\theta }_A}{g}>\frac{v_i{}^2\sin 2{\theta }_B}{g}$

Sine is increasing function between $0^o\text{to }{90}^o$

If above condition is satisfied, then we can say that $R_A>R_B$ otherwise can’t say anything.

One can’t say anything about which travels further.

Conclusion:

By comparison, angle of projection if initial velocities are same then we can compare maximum height, time of flight, but can’t compare range unless numerical values of angle of projection are known.