Question 4.4: That’s Quite an Arm! A stone is thrown from the top of a bui...
That’s Quite an Arm!
A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal with an initial speed of 20.0 m/s as shown in Figure 4.14. The height from which the stone is thrown is 45.0 m above the ground.
(A) How long does it take the stone to reach the ground?
(B) What is the speed of the stone just before it strikes the ground?
(A) Conceptualize Study Figure 4.14, in which we have indicated the trajectory and various parameters of the motion of the stone.
Categorize We categorize this problem as a projectile motion problem. The stone is modeled as a particle under constant acceleration in the y direction and a particle under constant velocity in the x direction.
Analyze We have the information x_i = y_i = 0, y_f = -45.0 m, a_y =-g, and v_i = 20.0 m/s (the numerical value of y_f is negative because we have chosen the point of the throw as the origin).
Find the initial x and y components of the stone’s velocity:
\begin{aligned}& v_{x i}=v_i \cos \theta_i=(20.0 m/s) \cos 30.0^{\circ}=17.3 m/s \\& v_{y i}=v_i \sin \theta_i=(20.0 m/s) \sin 30.0^{\circ}=10.0 m/s \end{aligned}Express the vertical position of the stone from the particle under constant acceleration model:
y_f=y_i+v_{y i} t-\frac{1}{2} g t^2Substitute numerical values:
-45.0 m=0+(10.0 m/s) t+\frac{1}{2}\left(-9.80 m/s^2\right) t^2Solve the quadratic equation for t:
t = 4.22 s
(B) Analyze Use the velocity equation in the particle under constant acceleration model to obtain the y component of the velocity of the stone just before it strikes the ground:
v_{y f}=v_{y i}-g tSubstitute numerical values, using t = 4.22 s:
v_{y f} =10.0 m/s+\left(-9.80 m/s^2\right)(4.22 s)=-31.3 m/sUse this component with the horizontal component v_{xf} = v_{xi} = 17.3 m/s to find the speed of the stone at t = 4.22 s:
v_f =\sqrt{v_{x f}{ }^2+v_{y f}{ }^2}=\sqrt{(17.3 m/s)^2+(-31.3 m/s)^2}=35.8 m/sFinalize Is it reasonable that the y component of the final velocity is negative? Is it reasonable that the final speed is larger than the initial speed of 20.0 m/s?