### Video Transcript

A body was projected vertically upwards at 53.9 metres per second. Given that at a certain time π‘ its height was 49 metres, find all the possible values of π‘. Take π equals 9.8 metres per second squared.

In this question, weβre given that a body or an object is projected upwards. Weβre asked to find all the possible values of π‘ when the height is 49 metres. We might wonder why thereβs more than one possible value of π‘. So letβs have a look at the path of this object. Weβre told that itβs projected upwards with a velocity of 53.9 metres per second. But at some point, the velocity will reach zero. And the body will return to the starting point. Therefore, our body could potentially be at a height of 49 metres on two separate times.

To answer this question, weβre going to use the kinematic equations of motion. So letβs start by seeing if we can assign values to any of these variables. Weβre asked to find the value of time π‘. So that variable will need to be in the equation of motion that we select. Weβre told that the body is projected upwards at 53.9 metres per second. So our π£ sub zero β thatβs our initial velocity β is 53.9 metres per second. You may also know initial velocity written with the letter π’ instead.

The height or displacement of this object is at 49 metres. So we can use the letter π to indicate this variable although it can often be seen with the letter π . And finally, our acceleration π is taken as negative gravity, which we can write as negative 9.8 metres per second squared. An equation of motion including these four variables will be π equals π£ sub zero π‘ plus a half ππ‘ squared. In terms of the letters π’ and π , the equation would be π equals π’π‘ plus a half ππ‘ squared. Substituting the values of our variables into this equation then will give us 49 equals 53.9π‘ plus a half times negative 9.8π‘ squared. Simplifying our right-hand side then, we have 49 equals 53.9π‘ minus 4.9π‘ squared.

Noticing that we now have a term in π‘ and a term in π‘ squared, weβre going to rearrange this quadratic into a form where it can easily be factored. So adding 4.9π‘ squared and subtracting 53.9π‘ from both sides of the equation will give us 4.9π‘ squared subtract 53.9π‘ plus 49 equals zero. Now we have our quadratic in a form where itβs easier to factor. But letβs make life a little bit easier by removing the decimals. And we can do this by multiplying our entire equation by 10.

We now have quite a few large coefficients. But we notice that we have 49 and 490. So letβs see if we can take out 49 as a common factor. Since 539 is also divisible by 49, then we can write our equation as π‘ squared minus 11π‘ plus 10 equals zero. So now, to factor our equation, weβre looking for two values which multiply to give 10 and add to give negative 11. Since the values of negative 10 and negative one would fit, then our factored form is π‘ minus 10 π‘ minus one equals zero.

To find a solution for π‘ then, we put each of the expressions in parentheses equal to zero and solve, giving us that π‘ minus 10 equals zero or π‘ minus one equals zero. In our first option then, when π‘ minus 10 is zero, then π‘ must be equal to 10 seconds. And in our second option, when π‘ minus one equals zero, then π‘ is equal to one second.

So our final answer for our time π‘ then is π‘ equals one second or π‘ equals 10 seconds.