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`
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?
1 answer
1 answer