A particle is projected vertically upwards at seven metres per second from a point 38.7 metres above the ground. Find the maximum height that the particle can reach. Consider the acceleration due to gravity to be 9.8 metres per second squared.
In order to solve this problem, we will use one of the equations of motion: 𝑣 squared equals 𝑢 squared plus two 𝑎 𝑠, where 𝑣 is the final velocity, 𝑢 is the initial velocity, 𝑎 is the acceleration, and 𝑠 is the displacement. At the maximum height, our velocity is equal to zero metres per second. As the particle has been projected vertically upwards, the acceleration is negative 9.8. This gives us values of 𝑢 of seven, 𝑣 of zero, and 𝑎 of negative 9.8.
Substituting these into our equation, 𝑣 squared equals 𝑢 squared plus two 𝑎 𝑠 gives us zero equals 49 minus 19.6𝑠. Rearranging this equation gives 19.6𝑠 is equal to 49. Dividing both sides of this equation by 19.6 gives us 𝑠 equals 2.5. Therefore, the maximum height above the point of projection is 2.5 metres.
However, a particle started 38.7 metres above the ground. Therefore, to work out the maximum height, we need to add 38.7 and 2.5. This gives us an answer of 41.2 metres. Therefore, the maximum height the particle can reach is 41.2 metres above the ground.