Question Video: Studying the Motion of a Particle Projected Vertically Upward from a Given Height given the Distance It Covers during a Given Second Mathematics

Video Transcript

A ball is projected vertically upward from the edge of a cliff that is 186.2 meters above the ground. If, during the third second, it covers a distance of 2.45 meters upward, find the time that the ball takes to reach the ground at the bottom of the cliff. Take the gravitational acceleration to be 9.8 meters per square second.

When dealing with a problem of this kind, it is often useful to draw a diagram first. We know that the cliff is 186.2 meters high. A ball is projected vertically upwards as shown. We’re told that acceleration is 9.8 meters per square second. As the ball is projected upwards, for the purpose of this question, we will take 𝑎 to be negative 9.8. Gravity is working against the direction of the projection.

In order to answer our question, we will use our equations of motion, often known as the SUVAT equations. We are told that during the third second, the ball covers a distance of 2.45 meters upward. We will call the velocities at the start and end of this second 𝑣 one and 𝑣 two. Focusing just on this part of the projection, 𝑢 is equal to 𝑣 one and 𝑣 is equal to 𝑣 two. These are the initial and final velocities. The displacement 𝑠 is 2.45 meters. Acceleration 𝑎 is negative 9.8. As this occurred in a one-second period, 𝑡 is equal to one.

In order to calculate the value of 𝑣 one, we can use the equation 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values, we have 2.45 is equal to 𝑣 one multiplied by one plus one-half multiplied by negative 9.8 multiplied by one squared. The right-hand side simplifies to 𝑣 one minus 4.9. Adding 4.9 to both sides of this equation gives us 𝑣 one is equal to 7.35. The velocity at the start of the third second is 7.35 meters per second.

We can now use this value to calculate the initial velocity 𝑢 of the ball. This deals with the first two seconds of the projection; therefore, 𝑡 is equal to two. 𝑣, the final velocity of this part, is 7.35. Once again, 𝑎 is equal to negative 9.8. This time, we will use the equation 𝑣 equals 𝑢 plus 𝑎𝑡. Substituting in our values gives us 7.35 is equal to 𝑢 plus negative 9.8 multiplied by two. Negative 9.8 multiplied by two is negative 19.6. Adding 19.6 to both sides of this equation gives us 𝑢 is equal to 26.95. The initial speed that the ball is projected at is 26.95 meters per second.

We are now in a position to find the time that the ball takes to reach the ground. Our value of 𝑢 is 26.95. Once again, 𝑎 is negative 9.8. The displacement is negative 186.2; this is the height of the cliff. As the ball was projected vertically upwards, the displacement will be negative. In order to calculate our value of 𝑡, we will use the equation 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values gives us negative 186.2 is equal to 26.95𝑡 plus a half multiplied by negative 9.8𝑡 squared. The final term simplifies to negative 4.9𝑡 squared.

We now have a quadratic equation that we need to set equal to zero. We do this by adding 4.9𝑡 squared and subtracting 26.95𝑡 from both sides. We could solve this quadratic equation using the quadratic formula. Alternatively, we might notice that all three terms are divisible by 4.9. The equation simplifies to 𝑡 squared minus 5.5𝑡 minus 38 is equal to zero. Once again, we could use the quadratic formula to solve this.

Alternatively, we could factorize it into two brackets, 𝑡 minus 9.5 and 𝑡 plus four. This gives us two possible values of 𝑡 of 9.5 and negative four. As time must be positive, we can rule out negative four. We can therefore conclude that the time taken for the ball to reach the ground at the bottom of the cliff is 9.5 seconds.