A rock dropped from a bridge takes 3.21 seconds to hit the water. How high is the bridge in meters?
I need to find Vsub0 to be able to calculate h....but how do I get Vsub0???
(1/2)(9.81)(3.21)^2 = 50.5 meters
more formally
h = Hi + Vi t - 4.9 t^2
0 = Hi + 0 - 4.9 t^2
Hi = 4.9 *3.21^2 = 50.5
the initial speed is zero. It was dropped, not thrown.
yes, good
is it -9.80 x 3.21= -31.458 is Vsub0??
h=151.47m??
accck ok so
1/2(9.80)(3.21)^2= 50.5 m
I'm learning slowly but surely!
To calculate the height of the bridge, you need to find the initial velocity (V0) of the rock when it was dropped. You can use the formula for free fall:
h = V0 * t + (1/2) * g * t^2
Where:
h = height of the bridge
V0 = initial velocity of the rock
t = time taken for the rock to hit the water (given as 3.21 seconds)
g = acceleration due to gravity (approximately 9.8 m/s²)
Since the rock was dropped from rest (meaning its initial velocity was 0), the equation simplifies to:
h = (1/2) * g * t^2
Now you need to substitute the known values into the equation:
h = (1/2) * (9.8 m/s²) * (3.21 s)^2
Simplifying:
h = (1/2) * 9.8 m/s² * 10.3041 s²
h ≈ 50.97299 m
Hence, the height of the bridge is approximately 50.97 meters.