A car accidently rolls off a cliff.as it leaves the cliff has a horizontal velocity of 113m/s it hits the ground 60m from the shoerline calculate the height off the cliff
it was in the air 60/113 seconds.
how high was it? hi=1/2 g t^2. solve for hi
Height of the cliff
Physics
To calculate the height of the cliff, we can use a kinematic equation that relates the horizontal and vertical components of motion.
Let's assume that the initial height of the car above the shore is "h" meters. We know that the car has a horizontal velocity of 113 m/s and hits the ground 60 m from the shoreline.
Using the kinematic equation:
\[d = v_{0x} \cdot t\]
where:
d = horizontal distance traveled (60 m)
v_{0x} = initial horizontal velocity (113 m/s)
t = time taken to reach the ground
The horizontal component of motion remains constant throughout. Therefore, we can rewrite the formula as:
\[60m = 113m/s \cdot t\]
Solving for t, we get:
\[t = \frac{60m}{113m/s} \approx 0.5309s\]
We can now use another kinematic equation to find the height (h) of the cliff:
\[h = v_{0y} \cdot t + \frac{1}{2} \cdot g \cdot t^2\]
where:
v_{0y} = initial vertical velocity (unknown)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach the ground (0.5309 s)
Since the car falls freely, its initial vertical velocity is 0 m/s (as it starts from rest). We can now substitute the values into the formula:
\[h = 0 \cdot 0.5309s + \frac{1}{2} \cdot (-9.8 m/s^2) \cdot (0.5309s)^2\]
Simplifying the equation:
\[h = -4.9 \cdot (0.5309s)^2\]
Calculating h, we obtain:
\[h \approx 0.6969m\]
Therefore, the height of the cliff is approximately 0.6969 meters.