Chapter 5, Problem 30A

A projectile is launched vertically at 50 m/s. If air resistance can be neglected, at what speed will it return to its initial level? Where in its trajectory will it have minimum speed?

To determine

The speed of the projectile while returning to its initial position. Also, the minimum speed location in its trajectory.

Answer to Problem 30A

Projectile came to its initial position with 49.99 m/s speed and the speed is minimum at the top of height.

Explanation of Solution

Given:

Initial velocity of the projectile along vertical direction is $v_y\text{= 50 m/s}$

Projectile is projected vertically. Hence the angle of projection is $\theta \text{=}\text{90}°$

Formula:

Velocity of the projectile along vertical direction is,

$v_y\text{= }u\text{sin}\left(\theta \right)$

Position where projectile has minimum speed is maximum height of the projectile is,

  $H=\frac{u^2{\sin }^2\left(\theta \right)}{2g}$

Where,

g = Acceleration due to gravity = 9.8 m/s²

Speed of the projectile when it comes to its initial position can be found by using kinematics equation,

  $v_f{}^2=v_i{}^2+2aH$

Calculation:

Speed of the projectile when it comes to its initial position is,

  $\begin{array}{l}{v}_y=u\sin \left(\theta \right)\\ u=\frac{50\text{m/s}}{\sin \left(90°\right)}\\ u=\text{50}\text{m/s}\end{array}$

Position where projectile has minimum speed is maximum height of the projectile. So, the maximum height will be,

  $\begin{array}{l}H=\frac{u^2{\sin }^2\left(\theta \right)}{2g}\\ H=\frac{{\left(50\text{m/s}\right)}^2{\sin }^2\left({90}^0\right)}{2\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ H=127.55\text{m}\end{array}$

Velocity at which projectile hits the ground,

  $\begin{array}{l}{v}_f{}^2=v_i{}^2+2aH\\ \text{Since}\text{at}\text{height}H,\\ v=0\text{m/s}\\ v_f{}^2=v_i{}^2+2aH\\ v_f{}^2={\left(0\text{m/s}\right)}^2+2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(127.55\text{m}\right)\\ v_f{}^2=2499.98\\ v_f=\text{49}\text{.99}\text{m/s}\end{array}$

Conclusion:

Projectile came to its initial position with 49.99 m/s speed and it is at distance of 127.55 m vertically where it has minimum speed.