Initially, velocity of the stone,u = 5 m/s Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height) Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2 (in downward direction) There will be a change in the sign of acceleration because the stone is being thrown upwards. Acceleration, a = −10 m/s2 Let s be the maximum height attained by the stone in time t. According to the first equation of motion: v = u + at 0 = 5 + (−10) t `thereforet=(-5)/-10=0.5s` According to the third equation of motion: v2 = u2 + 2 as (0)2 = (5)2 + 2(−10) s `s=5^2/20=1.25m` Hence, the stone attains a height of 1.25 m in 0.5 s. (Question 20 in AQA calculator paper from November 2017)This question is really wordy, so the first thing we want to do is condense all the information we are given into more manageable equations! The first important piece of information the question tells us is that 'h is directly proportional to v2' . This sounds scary, but all it means is that the height increases by a constant multiple for every increase in the value of (velocity)2. Instead of complicating this with words, we introduce the constant multiple k and create an equation that we can use to work out the answer.Step 1/ Create an equation that will look like this:height= k (velocity)2 or h= kv2 Step 2/ We now need to work out the value of k. We can do this by plugging in the values that we are given (when the stone is thrown at 10m/s, the maximum height reached is 5metres, therefore h=5 when v=10)(5)= k(10)2expand to get : 5=100ktherefore: k=0.05 (or 1/20)Step 3/ We now have to work out what the max height is when the velocity is 24. We can do this by plugging 24 back into the initial equation, where the value of k is 0.05:h=0.05v2 h=0.05(24)2 then use your calculator to get: h=28.8metres and you're done!
10 Questions 10 Marks 10 Mins
CONCEPT:
v = u + at \(s= ut +\frac{1}{2}at^{2}\) CALCULATION: Given: v = 0 (since the velocity will become zero at the highest point) u = 60m/s a = -10m/s (acceleration due to gravity and "-" sign for the opposite direction. using the the formula: v = u + at 0 = 60 -10t t = 6 sec.
\(s= ut +\frac{1}{2}at^{2}\) \(s = 60 \times 6 + \frac{1}{2}\times -10\times 6 \times 6\) \(s = 180m\)
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What is the maximum height a stone will reach if it is thrown upward with a velocity of 2 Metre per second?
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