Exciting Physics Problem: Stunt Car Freefall Calculation!
How far does a movie stunt car travel during its fall off a cliff?
Ignores air resistance, a stunt car is driven off a 44.1m-high cliff with a horizontal speed of 10.0 m/s. Given the gravitational constant g=9.8 m/s², how far does the car travel during its fall?
Group of answer choices:
a. 40 m
b. 30 m
c. 50 m
d. 60 m
Final answer:
The stunt car will travel horizontally a distance of 30 meters during its fall. How is this result determined?
Answer:
The stunt car will travel horizontally a distance of 30 meters during its fall from the cliff. This calculation involves analyzing the horizontal and vertical motion of the car, determining the time in freefall, and using the constant horizontal speed to find the total horizontal distance traveled.
The motion of the stunt car during the fall can be broken down into two components: horizontal travel and falling under gravity. The horizontal motion maintains a constant speed equivalent to the horizontal speed of the car, unaffected by the vertical freefall. The vertical motion is influenced solely by gravity's constant acceleration.
To find the horizontal distance traveled by the car, we first determine the fall time using the formula: t = sqrt(2h/g), where 'h' denotes the height and 'g' represents the acceleration due to gravity. Plugging in the values, we calculate: t = sqrt((2 * 44.1m)/(9.8m/s²)) ≈ 3.0s.
Using the equation: x = vt, with 'v' as the horizontal speed and 't' as the fall time, we can calculate the horizontal distance covered: x = 10.0m/s * 3.0s = 30m. Therefore, the correct answer is b. 30m.
Explore more about Projectile Motion for further insights and interesting physics concepts!