Question Video: Finding the Height an Upward Projected Body Reaches after a Given Time Mathematics

A body was projected vertically upwards at 18.3 m/s from a point 163 m above the ground. Find the position of the body 5 seconds after it was projected. Take 𝑔 = 9.8 m/s².

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Video Transcript

A body was projected vertically upwards at 18.3 meters per second from a point 163 meters above the ground. Find the position of the body five seconds after it was projected. Take 𝑔 equal to 9.8 meters per square second.

We are told that a body is projected vertically upwards with a velocity of 18.3 meters per second. It begins 163 meters above the ground. We are also told that the acceleration due to gravity is 9.8 meters per square second. This acts in the downward direction.

In order to solve this problem, we’ll use our equations of motion or SUVAT equations. We are interested where the body is five seconds after it was projected. This means that the time 𝑡 is equal to five. The initial velocity 𝑢 is 18.3. 𝑎 is equal to negative 9.8 as the body was projected vertically upwards. We will let the displacement after five seconds be 𝑥.

The displacement is the distance away from the origin or point of projection. This could be above or below the start point. If our value of 𝑥 is negative, it will be below the start point, whereas if 𝑥 is positive, it will be above the point of projection. In order to calculate the value of 𝑥, we will use the equation 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values, we have 𝑥 is equal to 18.3 multiplied by five plus a half multiplied by negative 9.8 multiplied by five squared. The right-hand side simplifies to 91.5 minus 122.5. 𝑥 is equal to negative 31.

This means that after five seconds, the body is 31 meters below the point of projection. 163 minus 31 is equal to 132. We can therefore conclude that after five seconds, the body is 132 meters above the ground.

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