Mr. Tiet is driving her car at a speed of 55 km/h. he accidentally drops the phone out the window. he drops the phone from 4.0 feet.
a)How far did the car travel when the phone hit the ground?
b)What was the final velocity?
well, we will have to chose metric or English so let's say we drop it from 1.22 meters
v = - g t = -9.81 t
h = 1.22 - (1/2)(9.81) t^2 = 0 when it hits ground
so
t^2 = 1.22/4.9
t = 0.50 seconds
so v = -9.81 * .5 = - 4.9 meters/second
car going 55,000 meters / 3600 seconds = 15.3 meters/second
so the car goes 7.65 meters during that half second
the final velocity of the phone is 4.9 m/s down and 15.3 m/s horizontal
magnitude = sqrt (15.3^2 + 4.9^2)
Given: V = 55 km/h = 15.3 m/s. h = 4 Ft. = 1.21 m.
a. 0.5*g*T^2 = 1.21
0.5*9.8*T^2 = 1.21
T = 0.5 s. to hit gnd.
d = V*T = 15.3 * 0.5 = 7.65 m.
b. Y = Yo + g*T = 0 + 9.8*0.5 = 4.9 m/s.
V = Xo + Yi = 15.3 + 4.9i
V = sqrt(15.3^2 + 4.9^2) =
To answer these questions, we need to consider the concept of projectile motion. When an object is dropped from a moving vehicle, it will have both vertical and horizontal components of motion.
a) To determine the horizontal distance the car traveled when the phone hit the ground, we need to find the time it took for the phone to fall. Assuming there is no air resistance, the time would be the same for both horizontal and vertical motions. We can use the formula:
d = v * t
where "d" is the horizontal distance, "v" is the horizontal velocity (which is the same as the car's velocity), and "t" is the time.
To find the time, we can use the vertical motion component. The phone was dropped from a height of 4.0 feet, which is equivalent to 1.22 meters. We can use the equation of motion to calculate the time it takes to fall:
d = (1/2) * g * t^2
where "d" is the distance (1.22 meters), "g" is the acceleration due to gravity (approximately 9.8 m/s^2), and "t" is the time.
Rearranging the equation gives:
t^2 = (2 * d) / g
Substituting the values, we find:
t^2 = (2 * 1.22) / 9.8
t^2 = 0.25
t = 0.5 seconds
Since the time is the same for both horizontal and vertical motions, we can use this value to find the horizontal distance:
d = v * t
d = 55 km/h * (0.5 hours * 3600 seconds/hour)
d = 55 * 0.5 * 3600 / 1000
d = 99 meters
Therefore, the car traveled approximately 99 meters when the phone hit the ground.
b) To find the final velocity, we need to determine the horizontal and vertical velocities separately. The horizontal velocity remains constant at 55 km/h as there are no horizontal forces acting on the phone. The vertical velocity is influenced by gravity and will increase as the phone falls.
The final vertical velocity just before the phone hits the ground can be found using the equation:
v = u + g * t
where "v" is the final velocity, "u" is the initial vertical velocity (which is 0 m/s since the phone was dropped), "g" is the acceleration due to gravity (approximately 9.8 m/s^2), and "t" is the time (0.5 seconds).
Substituting the values, we have:
v = 0 + 9.8 * 0.5
v = 4.9 m/s
Therefore, the final velocity of the phone just before it hits the ground is 4.9 m/s.