Physics - cell phone word question

Question

A person accidently throws his cellphone out of an open window with a velocity of 4.5 m/s at an angle 25 ̊above the horizontal. The student looks out the window and notices a friend outside. The window is above the ground. How far away from the building should the friend stand to catch the cell phone at ground level?

Answer 1

If both the thrower and the catcher are at the same level (ground level), then a sketch might look like this:

We can break the velocity vector into it's two components, v_x and v_y. Using trig, we can say:

cos(25°) = v_x/4.5 or v_x = 4.5cos(25°) = 4.078 m/s

sin(25°) = v_y/4.5 or v_y = 4.5sin(25°) = 1.902 m/s

To find the time of the flight, let's look ONLY in the y-direction. Knowing that, in the y-direction, the velocity at the top of the flight, in the y-direction, is zero, we can say (using the kinematic equations of motion):

v_f = v_i + at where v_f = 0 m/s, v_i = 1.902 m/s, a = g = -9.81 m/s² and t = time in seconds:

0 = 1.902 + (-9.81)t

9.81t = 1.902

t = 0.1939 seconds.

But to find the full time the cellphone is in the air, we need to double the time (it take 0.1939 sec to reach the top of its flight). So total time = (0.1939)2 = 0.3877 seconds.

During that 0.3877 seconds, the cellphone is also traveling in the x-direction. The distance it travels is its speed times its time.

x = v_xt

x = (4.078 m/s)(0.3877 seconds) = 1.58 meters

So the "catcher" needs to be 1.58 meters away from the "thrower".