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2. If I throw a ball in the air at 13 m/s, what is the height of the cliff I am on if it takes 4 seconds for it to hit the ground?

User Margi
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1 Answer

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Final answer:

Using the kinematic equation and the given initial velocity and time, the height of the cliff from which the ball is thrown is calculated to be approximately 52.96 m.

Step-by-step explanation:

To determine the height of the cliff from which the ball is thrown, we can use the following kinematic equation for uniformly accelerated motion (considering the acceleration due to gravity):

y = v0t + ½at2

Here, y is the displacement (height of the cliff), v0 is the initial velocity (13 m/s upwards), t is the time of flight (4 seconds), and a is the acceleration due to gravity (-9.81 m/s2, negative because it is directed downwards).

When substituting the values into the equation:

y = (13 m/s)(4 s) + ½(-9.81 m/s2)(4 s)2

y = 52 m - 78.48 m

y = -26.48 m

Because the displacement is negative, it means the ball landed below the height from which it was thrown. However, since we know the ball takes 4 seconds to reach the ground, the first 1.326 seconds is spent climbing to its highest point, and the remaining time is spent falling from that point to the ground. Thus, we must double the calculated height (as it falls the same distance it climbed).

The height of the cliff is therefore:

y = -2 × (-26.48 m) = 52.96 m

User Lyngvi
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