A girl standing on a bridge throws a stone vertically downward with an intial velocity of 15.0 m/s into the river below. If the stone hit's the water 2.00 seconds later what is the height of the bridge above the water?
h=vi*t+1/2 a t^2
h=-15*2-4.9 *4
h=-30-19.6=-49.6m to the water, so the height of the bridge is 49.6 above the water.
physics Kid: Nope, it is not correct.
To determine the height of the bridge above the water, we can use the kinematic equation for vertical motion:
h = h0 + v0t + (1/2)gt^2
where h is the final height, h0 is the initial height (in this case, the height of the bridge), v0 is the initial velocity, t is the time of flight, and g is the acceleration due to gravity.
Given:
v0 = 15.0 m/s (downward)
t = 2.00 s
g = -9.8 m/s^2 (taking downward as negative)
Substituting these values into the equation, we can solve for h:
h = 0 + (15.0 m/s)(2.00 s) + (1/2)(-9.8 m/s^2)(2.00 s)^2
h = 0 + 30.0 m - 19.6 m
h = 10.4 m
Therefore, the height of the bridge above the water is 10.4 meters.
Everything up is positive
Everything Down is negaitive
velocity initial = -15.0m/s
distance final = 0m
distance initial = ?
time = 2s
df=v * t+di
0 = -15 * (2)+di
di = 30 m
therefor the bridge is 30m ablive the water.
Not sure if this is entirely correct.