A body fell vertically from the top of a tower. It covered 86.73 meters in the final second before hitting the ground. Determine the height of the tower rounding your answer to the nearest two decimal places. Let the acceleration due to gravity 𝑔 equal 9.8 meters per square second.
We are told that a body fell from the top of a tower. This means that its initial velocity 𝑢 is equal to zero meters per second. In the last second, it covered a distance of 86.73 meters. We will let the velocity at which it hit the ground equal 𝑣. We are also told that the acceleration due to gravity is 9.8 meters per square second. We need to calculate the height of the tower labeled 𝑥. We will do this using the equations of motion or SUVAT equations.
In the final second of motion, the displacement 𝑠 is 86.73, 𝑎 is equal to 9.8, and 𝑡 is equal to one. We can use the equation 𝑠 is equal to 𝑣𝑡 minus a half 𝑎𝑡 squared to calculate the final velocity 𝑣. Substituting in our values, we have 86.73 is equal to 𝑣 multiplied by one minus a half multiplied by 9.8 multiplied by one squared. The right-hand side simplifies to 𝑣 minus 4.9. Adding 4.9 to both sides of this equation gives us a value of 𝑣 of 91.63. The velocity of the body when it hits the ground is 91.63 meters per second.
We’ll now use this information to calculate the height of the tower. We recall that the initial velocity 𝑢 is equal to zero, 𝑣 is equal to 91.63, 𝑎 is equal to 9.8. We can use the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠 to calculate the height of the tower. Substituting in our values gives us 91.63 squared is equal to zero squared plus two multiplied by 9.8 multiplied by 𝑥.
91.63 squared is equal to 8396.0569. The right-hand side simplifies to 19.6𝑥. Dividing both sides by 19.6 gives us 𝑥 is equal to 428.37025. We’re asked to round our answer to two decimal places. We can therefore conclude that the height of the tower is 428.37 meters.