Hey Chris, this is regarding the last question. The stone takes 0.8s to reach max height, it would take another 0.8s to go back to its original release point which means it takes 1.6s for its journey from the hand to max height and back to the hand, now since we are not taking the height between the hand and the foot of cliff the into consideration, it theoretically means that the stone is being released from the foot of the cliff. Now to calculate the height of the cliff, we have to consider finding the height between the foot of the cliff and the sea level. This means that the time taken from the base of the cliff to reach the sea level will have to be 3s - 1.6s = 1.4s (since the entire journey of the stone is 3s, it means it goes from the foot of the cliff to the max height and then to the sea in those 3s, and since we have to find the height of the cliff only, we'll have to subtract the journey of the stone reaching max height and back to its origin which is 1.6s). This means the actual height of the cliff would be 1.4 m ( 8(1.4)+(-5)(1.4)^2). Please let me know if im wrong.
according to question 3 of May 2015 TZ1 physics SL paper 1, it says, when a tennis ball is released from rest and falls vertically through a small distance in air, the acceleration of the ball decreases. Is this because of air resistance? (Cuz I thought we always consider air resistance as negligible in all ib questions)
at the minute 20:52 the question says there is 3 seconds between the ball leaving antonia's hand and hitting the sea and do not forget that it took 0.8 seconds for the ball to reach the max height so the time neeeded for the ball to get from the max height is 2.2. Should not this be substituted insted of the the three in the equation of x=ut - 5tt ?
The nice thing about the equation is it tells us the height after any elapsed time, whether it is on the way up or down. If you want to make this a complicated problem, you could take t=0 to be the top of the flight, and then use 2.2 seconds to find the height relative to the maximum height.
Another way to think about it is this: It takes 0.8 s to reach the max height and an EXTRA 0.8 sec for the rock to reach the height of Antonia. That is a total of 1.6 seconds. However, it takes 3.0 seconds for the rock to leave Antonia's hand, reach the max height, come back down to Antonia's height, then reach the bottom of the cliff. It takes 3 seconds to do all of that. Thus, ON THE WAY DOWN, the time it take for the stone to travel from Antonia's height(on its way DOWN) to the bottom of the cliff is 3.0 - 1.6 seconds = 1.4 seconds. Then substitute that in the kinematic equation with 8.0 m/s for the initial velocity and you get 21 m. I think the mistake you made was that you misread the question. I did the same at first, but then I fixed it.
Isn't average speed supposed to be total distance/ total time. I'm asking this because I was told only to use this particular formula many times and not to just take the average of the speeds because the answer would be incorrect sometimes
Right you can not take the average of the speeds unless the speeds were carried out for the same amount of time, but you can always use total distance over time.
Sir, I don't understand for the chart question, why the equation for the height is 30t^2-5t^2. Also, why do we use the initial velocity? Shouldnt we use the average velocity? Thanks!
+Ehtesham B That is 30t - 5t^2. The first term, 30t, represents how far the object would go if there were no gravity and it moved with constant speed (i.e. dist=speed x time). The second term is due to gravity acting on the object, and contains a t^2 because the speed is changing. For any object starting from rest, with acceleration a, the distance traveled will be given by 1/2 a t^2.
Sir, for the warmup questions, why isn't the average speed for the 3 seconds 20 meters/second? ( 10m + 20m + 30m). Thank you so much for making these videos. They are very helpfull.
+Ehtesham B It only goes 5m in the first second not 10m. It increases its speed from 0 to 10 m/s over the first second because its acceleration is 10m/s every second. The average speed will then be half of 10 or 5m/s over the first second. If the object falls for 1 s at an average speed of 5 m/s it will fall 5. The reasoning is exactly the same for the 3 second interval.
Since they are asking for height of cliff. Shouldnt you minus the height from the student to the maximum (3.2) and minus that from the overall height (21m) and get the height of the cliff
If you scored 0 on one test and 30 on another test, would your average be 15 or 10? If the object had travelled 30 m in 3 seconds, the average speed would be 10, but it is speeds that we know not distances.
@@donerphysics But you still are taking into account only 2 tests. In the case of the warmup question, the tests should be taken as 3 as there are 3 seconds. So if it is 3 tests, the denominator will be three. So the answer with (0+30)/3. pls correct if I'm wrong
