Question

A physics teacher walks at a very constant and amazing velocity of 2 m/s. A sneaky and disgruntled physics student stands on the top of a 20 m tall building and plans to drop the balloon on the head of the caring and genuine 2 m tall physics teacher below. How far away should the amazing teacher be from the impact point when the student releases the balloon?

a) 10 m
b) 20 m
c) 30 m
d) 40 m

Answer 1

To find the distance at which the physics teacher should stand from the point where the balloon will be dropped, we can use the equations of projectile motion.

Step-by-step explanation:

To find the distance at which the physics teacher should stand from the point where the balloon will be dropped, we can use the equations of projectile motion. The vertical motion of the balloon can be treated as free fall, while the horizontal motion is at a constant velocity of 2 m/s.

First, we need to find the time it takes for the balloon to fall from the top of the building to the ground. Using the equation h = (1/2)gt^2, where h is the height of the building (20 m) and g is the acceleration due to gravity (9.8 m/s^2), we can solve for t.

t = sqrt(2h/g) = sqrt(2 * 20 / 9.8) = 2.02 s

Next, we can find the horizontal distance traveled by the teacher during this time. Using the equation d = vt, where v is the horizontal velocity (2 m/s) and t is the time (2.02 s), we can solve for d.

d = 2 * 2.02 = 4.04 m

Therefore, the teacher should stand approximately 4.04 meters away from the impact point when the student releases the balloon. The correct answer is (a) 10 m.