Thank You so much for making this video. It really helped me in my online class, especially self study is not easy. Thank you for sharing your ideas that is very clear and so understandable. God bless and more power!
The video is really helpful, thank you. But I feel like it would have been more clear if you used numbers on your examples, and can the final range formula be further simplified into (v initial^2*(sin teta*)2)/g because I assume the angles will be the same?
If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-HPehCUv6bEY.html&ab_channel=Math%2CPhysics%2CEngineering
Thanks alot sir, excellent way of explaining complicated stuff... I had huge doubts in projectile motion, i looked at many videos but only found this one useful and best...
my physics exam is tomorrow and i still dont understand projectiles which will most definitely be there but theres so many other units to study theres no time 😭
Really wish I could hear this video. Turned my TV volume all the way up on the chrome cast, tried AirPods, the sound is so low. I do appreciate the making of it tho! Just a remark
Thank you so much! This video really helped me see the breakdown of the equations used to find the specifics (time, max height, vertical height, etc.) Please keep making videos and thank you :)
For anybody needing maximum height you need to square the answer to velocity x (sin) angle and then divide by 2 x 9.8. This video makes a bit of a pigs ear of explaining that
Which time(s) and distance(s) are you given? If you have the ∆x for t(max), or position & time of impact, then you should have your Vx for when Vy = 0 because Vx is constant. If your starting height and impact height are the same, the parabola from start to finish will be symmetrical, therefore, its midway point will be half of t(max). Long story short, you have to work backwards, and your max height is tangeant to the highest point on the parabola.
If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-HPehCUv6bEY.html&ab_channel=Math%2CPhysics%2CEngineering
what if the projectile had acceleration, per say, a rocket. the rocket acceleration is in a changing direction. and the parabola is less circular (longer on the +y path and shorter on the -y). What would be added?
this equation is true for a cannon or snipers sake, but a rocket that dosent have one blast that sets it in motion (rather a continuous acceleration in its trajectory ) is written differently, How?
how would i do a question lik this A projectile is fired with an initial velocity of 120 ms-1. The projectile has a time of flight of 18.75s. Determine: a) The range of the projectile; b) The maximum height attained, and the time at which this height is attained;
Great vid but i have a small question, in my school book it says that time to reach maximum height= -vy divided by g, which is -9.8m/s2, is there a difference between this and the one in the video?
at 14:26 i write an equation for t_top=v_0*sin(theta)/g. This is the same as the equation you wrote. the vy in your equation is the initial y component of the velocity.
If at a time T the direction of the velocity is at 90 degrees to the initial direction of the velocity is given in a problem what can you derive from that?
is the initial height same as the maximum height? cuz i have only have the velocity and the angle, ive already solved for the max height but the time and range i cant seem to solve it because i need the int. height
In linear motion velocity and acceleration are in the same direction but in the projectile motion velocity is upward and acceleration is downward how is that possible could you explain it to me.
I need help in calculating the trajectory length. Yes, it's the length of the parabola or projectile not the horziontal or vertical displacement. Any help?
You first need to write y vs x. You can do this by eliminating time. Then you will need to find the arc length. This is a standard math problem. I suggest googling it. It will involve integrating a square root function of dy/dx
In real life world, would it be fair to assume that there would be a deceleration of the motion in x-axis due to air friction? or is it negligible? If negligible, does that mean if i plot the graph of horizontal distance covered over time it will be a straight line and straight to zero when the final Y is 0?
In the real world, you would need to include air resistance as it has a non-negligible deceleration on the x and y motion of the projectile. I can’t remember the formula for drag off the top of my head but it is necessary for the real world.
