Question

A projectile is fired from the top of a 120 m tall building with an initial velocity of Vi = 72.0 m/s and an initial angle theta i =25.0° above the horizontal.Answer the following questions. [Use g= 9.80 m/s^2]Question 13What is the speed of the projectile the moment it touches the ground?Round your answer to 3 significant figures.

Answer 1

The vertical component of the speed changes uniformly with time as follows:

$v_y=v_{0y}-gt$

The horizontal component of the speed remains constant:

$v_x=v_{0x}$

To find the initial values for the components of the speed, remember the following:

$\begin{gathered} v_{0y}=v_0\sin (\theta) \\ v_{0x}=v_0\cos (\theta) \end{gathered}$

Substitute the values for v₀ and θ:

$\begin{gathered} v_{0y}=72.0\frac{m}{s}\times\sin (25º)=30.4\frac{m}{s} \\ v_{0x}=72.0\frac{m}{s}\times\cos (25º)=65.3\frac{m}{s} \end{gathered}$

Since the projectile touches the ground at time t=8.95s, use that value to find the final vertical speed:

$\begin{gathered} v_y=v_{0y}-gt \\ =(30.4\frac{m}{s})-(9.80\frac{m}{s^2})(8.95s) \\ =-57.3\frac{m}{s} \end{gathered}$

The final speed of the projectile when it touches the ground is given by:

$\begin{gathered} v=\sqrt[]{v^2_x+v^2_y} \\ =\sqrt[]{(65.3\frac{m}{s})^2+(-57.3\frac{m}{s})^2} \\ =86.8\frac{m}{s} \end{gathered}$

Therefore, the speed of the projectile the moment it touches the ground, is:

$86.8\frac{m}{s}$