a diver running 1.6m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.0s later. how high was the cliff and how far from its base did the driver hit the water?
Consider free fall:
h=1/2 g t^2 calculate h
distancehorizonal=initialhorizontalvelocity*timeinair
i got this 44.1m and 4.8m right?
yes, correct.
howd u get that?????
how did you get the height? how did you get 4.8 m???
isn't it 44.1 m and 5.4 ?
To find the height of the cliff and the horizontal distance from the base of the cliff to the point of impact in the water, we can use kinematic equations.
Let's break down the problem into two parts:
1. Finding the height of the cliff:
The height of the cliff can be found using the equation for vertical motion:
h = ut + (1/2)gt^2
In this case, the initial velocity in the vertical direction (u) is zero since the diver is running horizontally. The time (t) is given as 3.0 seconds, and the acceleration due to gravity (g) is approximately 9.8 m/s^2.
Substituting the given values:
h = 0 + (1/2)(9.8)(3.0)^2
Simplifying the equation gives:
h = 0 + (1/2)(9.8)(9.0)
h = 44.1 meters
Therefore, the height of the cliff is approximately 44.1 meters.
2. Finding the horizontal distance from the base:
The horizontal distance from the base of the cliff to the point of impact in the water can be found using the equation for horizontal motion:
s = ut
In this case, the horizontal velocity (u) is given as 1.6 m/s, and the time (t) is given as 3.0 seconds.
Substituting the given values:
s = 1.6 * 3.0
s = 4.8 meters
Therefore, the diver hits the water approximately 4.8 meters from the base of the cliff.
To summarize:
The height of the cliff is approximately 44.1 meters, and the diver hits the water approximately 4.8 meters from the base of the cliff.