# An object is launch horizontally at a velocity of 28 m/s from top of a cliff.If the object hit the ground 170 m from the base of the cliff, how high is cliff?

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To solve this problem, you need to be aware of certain aspects of ballistic physics, which in this case will be the physics described in the link below, but considering what happens if the angle is 0 degrees. The first assumption here is that there is no drag, which tells us that in the horizontal direction, the object is moving at a constant velocity. This gives us the following equation:

Dx = VxT

Here, Dx is the distance in the x-direction, Vx is the velocity in the x-direction, and T is the travel time.

Considering that travel in the vertical direction will be based on travel time, we will first need to sove for this variable, which is conveniently found with the information given in the problem. Substituting the above values into the equation:

170 = 28T

Dividing both sides by 28, we get our travel time:

T = 6.07 seconds

Now, we can forget everything about travelling in the horizontal direction because we are finding a vertical distance. This means, we don't care how about the horizontal velocities and distances, and we can simply treat this problem like one where you are dropping an object from a specific height! The relationship between distance in time in this case is given by the following equation:

Dy = 1/2gT^2

Here, Dy is the distance travelled in the y-direction, g is gravitational acceleration, and T is the total travel time. Here, g is a constant 9.8 m/s^2 and T, which we found above, will be 6.07 seconds. Thus, we can substitute these values into the equation to find Dy, our answer:

Dy = 1/2*9.8*6.07^2

**Dy = 180.5 meters**

We now have our answer. The cliff is 180.5 meters in height.

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