Question

A friend and another friend have a fight. She starts throwing his things out of the window , a distance of 46 m . What is the velocity of his prized bowling trophy as it hits the ground?

Answer 1

Using the principles of projectile motion and the equation v^2 = u^2 + 2gh, the velocity of the bowling trophy as it hits the ground after falling 46 meters is approximately 30.04 m/s.

Step-by-step explanation:

The question asks for the velocity of a bowling trophy as it hits the ground after being thrown out of a window. To find this velocity, we can use the principles of projectile motion and the acceleration due to gravity. Given that we are ignoring air resistance and the distance it falls is 46 meters, we can use the formula for the velocity of an object falling under gravity without any initial vertical velocity:

v^2 = u^2 + 2gh

where v is the final velocity, u is the initial velocity (which is 0 m/s straight down), g is the acceleration due to gravity (9.81 m/s^2), and h is the height (46 m). Plugging in the numbers:

v^2 = 0 + 2(9.81 m/s^2)(46 m)

v^2 = 902.52 m^2/s^2

v = √902.52 m^2/s^2

v ≈ 30.04 m/s

Therefore, the velocity of the bowling trophy as it hits the ground is approximately 30.04 m/s.