Chapter 3, Problem 14Q

A projectile is launched at an upward angle of 300 to the horizontal with a speed of 30 m/s. How does the horizontal component of its velocity 1.0 s after launch compare with its horizontal component of velocity 2.0 s after launch, ignoring air resistance? Explain.

Solution

To Compare: The horizontal components of velocity after 1 s and 2 s after launch.

Solution:

The horizontal component of velocity of projectile motion does not change during motion.

Explanation:

Given:

  

Initial velocity: $u=30\text{m/s}$

Launch angle: $\theta =30°$

Formula used:

Resolve projectile motion into two independent one-dimension motion, one is horizontal motion along with and another is vertical motion along.

Along horizontal direction, initial velocity is: $u_x=u\cos \theta$

Along vertical direction, initial velocity is: $u_y=u\sin \theta$

Along horizontal direction, acceleration is: $a_x=0$

Along vertical direction, acceleration is: $a_y=g$

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

Calculation:

First, calculate the time of flight:

  $\begin{array}{l}{T}_f=\frac{2u\sin \theta }{g}\\ T_f=\frac{2\times 30\times \sin {30}^{\text{o}}}{9.8}\\ T_f=3.06\text{s}\end{array}$

As there is no acceleration in the horizontal direction, therefore the horizontal component of velocity remains constant. Therefore, horizontal component at 1.0 and 2.0 s after launch is same.

Conclusion:

The horizontal component of velocity of projectile motion does not change during the time of flight (T).