A not so brilliant physics students wants to jump from their 3rd floor apartment window to the swimming pool below. The problem is the base of the apartment is 8.00 meters from the pool’s edge. If the window is 20.0 meters high, how fast does the student have to be running horizontally to make sure they make it to the pool’s edge?
We use the formula:
h = vo,y*t - (1/2)gt^2
where
h = height
vo,y = initial vertical velocity
t = time
g = acceleration due to gravity = 9.8 m/s^2
Since there is no vertical velocity involved, vo,y = 0.
If that's not very clear, we can say that vo,y = vo*sin(angle). But the student leaves the window horizontally (thus, angle = 0) and sin(0) = 0, so vo,y = 0.
Substituting,
20 = 0 - (1/2)(-9.8)(t^2)
20 = 4.9t^2
t^2 = 20/4.9
t = 2.02 s
Finally, use this equation for the range:
x = vo,x * t
where
x = horizontal distance
vo,x = initial horizontal velocity
t = time
8 = vo,x * 2.02
vo,x = 3.96 m/s^2
hope this helps~ `u`
To determine how fast the student needs to be running horizontally to reach the pool's edge, we can use the principles of projectile motion.
Given:
- The height of the window (h) = 20.0 meters
- The horizontal distance from the base of the apartment to the pool's edge (d) = 8.00 meters
Since the student jumps from a certain height and there are no initial vertical forces acting on the student, we can use the equation for free fall motion:
h = (1/2) * g * t^2
Where:
- h is the height of the window (20.0 meters)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time taken to fall
Simplifying the equation, we get:
t = √(2 * h / g)
t = √(2 * 20.0 / 9.8)
t ≈ √(40/9.8)
t ≈ √(4.08)
t ≈ 2.02 seconds
Now that we have the time it takes for the student to fall, we can calculate the horizontal velocity (v) required to reach the pool's edge:
v = d / t
v = 8.00 / 2.02
v ≈ 3.96 m/s
Therefore, in order for the student to make it to the pool's edge, they need to be running horizontally at a speed of approximately 3.96 meters per second.